RD Sharma Class 12 Exercise 15.3 Tangents and Normals Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 15.3 Tangents and Normals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 21, 2022 01:19 PM IST

Even though a teacher guides the students by teaching the concepts, there arises doubt when they try to solve an answer or a sum. This led to the need for high-quality solution books like the RD Sharma books. Especially while working out sums in maths, it is recommended to have a good set of solution books. RD Sharma solutions The widely used solution book for the students to understand the concepts is the RD Sharma Class 12th exercise 15.3 material.

## RD Sharma Class 12 Solutions Chapter 15 Tangents and Normals - Other Exercise

Chapter 15 - Tangents and Normals Ex 15.1

Chapter 15 - Tangents and Normals Ex 15.2

Chapter 15 -Tangents and Normals Ex-FBQ

Chapter 15 -Tangents and Normals Ex-MCQ

Chapter 15 -Tangents and Normals Ex-VSA

## Tangents and Normals Excercise:15.3

Tangents and Normals exercise 15.3 , question 1 sub question 1

$\theta=\tan ^{-1}\left(\frac{3}{4}\right) \text { and } \theta=\frac{\pi}{2}$
Hint – The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
m1=slope of first curve. m2=slope of second curve.
Given$-y^{2}=x \ldots \ldots(1) \& x^{2}=y \ldots \ldots(2)$
First curve is y2=x
Differentiating above with respect to x,
As we know, $\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=2 y \frac{d}{d x}=1 \\ &=m_{1}=\frac{d y}{d x}=\frac{1}{2 y} \ldots \ldots \text { (3) } \end{aligned}

Second curve is $x^{2}=y$

Differentiating above with respect to x,
\begin{aligned} &=2 x=\frac{d y}{d x} \\ &=m_{2}=\frac{d y}{d x}=2 x \ldots \ldots(4) \end{aligned}

Substituting (1) in (2),we get

\begin{aligned} &=x^{2}=y \\ &=\left(y^{2}\right)^{2}=y \\ &=y^{4}-y=0 \end{aligned}
\begin{aligned} &=y\left(y^{3}-1\right)=0 \\ &=y=0 \text { or } y^{3}-1=0 \\ &=y=0 \text { or } y^{3}=1 \\ &=y=0 \text { or } y=1 \end{aligned}
Substituting y=0 or y=1 in (1)
x=y2
When, y = 0, x = 0
y = 1, x = 1
Substituting the values of (y = 0, x = 0),(y = 1 , x = 1) for m1& m2 , we get,
When, y = 0
\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{1}{2(0)}=\frac{1}{0} \quad::\left\{\frac{1}{0}=\infty\right\} \\ &=m_{1}=\infty \end{aligned}

When, y = 1

$=m_{1}=\frac{d y}{d x}=\frac{1}{2(1)}=\frac{1}{2}$
Value of m1 is $\infty$ and $\frac{1}{2}$
When x = 0
\begin{aligned} &=m_{2}=\frac{d y}{d x}=2 x \\ &=m_{2}=\frac{d y}{d x}=2(0)=0 \end{aligned}
When x = 1
\begin{aligned} &=m_{2}=\frac{d y}{d x}=2 x \\ &=m_{2}=\frac{d y}{d x}=2(1)=2 \end{aligned}
Values of m2 is 0 and 2.
As we know, Angle of intersection of two curves is given by $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
When m1 is $\infty$ and m2 is 0
\begin{aligned} &\operatorname{tan} \theta=\left(\frac{\infty-0}{1+\infty \times 0}\right) \\ &\operatorname{tan} \theta=\infty \end{aligned}
$\theta=\tan ^{-1}(\infty)$ As we know $\tan \left(\frac{\pi}{2}\right)=\infty$
$\theta=\frac{\pi}{2}$
When $m_{1}=\frac{1}{2}$and $m_{2}=2$
\begin{aligned} &\operatorname{Tan} \theta=(1 / 2-2) /(1+1 / 2 \times 2)=\left(\frac{-3 / 2}{2}\right) \\ &\operatorname{Tan} \theta=\left(\frac{-3}{2} \times \frac{1}{2}\right)=\left(\frac{-3}{4}\right) \\ &=\theta=\tan ^{-1}\left(\frac{3}{4}\right) \end{aligned}

Tangents and Normals exercise 15.3 , question 1 sub question 2

$\theta=\tan ^{-1}\left(\frac{9}{2}\right)$
Hint – The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
m1=slope of first curve. m2=slope of second curve.
Given- $y=x^{2} \ldots \ldots(1) \;\; \& \;\; x^{2}+y^{2}=20 \ldots \ldots(2)$
First curve is $y=x^{2}$
Differentiating above with respect to x,
As we know, $\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=\frac{d y}{d x}=2 x \\ &=m_{1}=2 x \ldots \ldots(3) \end{aligned}
Second curve is $x^{2}+y^{2}=20$
Differentiating above with respect to x,
\begin{aligned} &=2 x+2 y \frac{d y}{d x}=0 \\ &=2 y \frac{d y}{d x}=-2 x \end{aligned}
\begin{aligned} &=m_{2}=\frac{d y}{d x}=\frac{-2 x}{2 y}=\frac{-x}{y} \\ &=m_{2}=\frac{-x}{y} \end{aligned}
Substituting (1) in (2),we get
\begin{aligned} &=y=x^{2} \\ &=y+y^{2}=20 \\ &=y^{2}+y-20=0 \end{aligned}
By factorization method
\begin{aligned} &=y^{2}+y-20=0 \\ &=y^{2}+5 y-4 y-20=0 \\ &=y(y+5)-4(y+5)=0 \\ &=(y-4)(y+5)=0 \end{aligned}
=y=-5 and y=4
Substituting y=-5 and y=4 in (1)
When, y = -5
$x^{2}=-5$ This is not possible
When y=4
$x^{2}=4\\ x=\pm 2$
Substituting the values for m1& m2 , we get,
\begin{aligned} &=m_{1}=\frac{d y}{d x}=2x \end{aligned}
When, x=2
$=m_{1}=2(2)=4$
When x=-2
$=m_{1}=2(-2)=-4$
Value of m1 is 4, -4
$=m_{2}=\frac{-x}{y}$
When y=4 & x=2
$=m_{2}=\frac{d y}{d x}=\frac{-x}{y}=\frac{-2}{4}=\frac{-1}{2}$
When y=4 & x=-2
$=m_{2}=\frac{d y}{d x}=\frac{-x}{y}=\frac{-(-2)}{4}=\frac{1}{2}$
Values of m2 is $\frac{1}{2}$ and $\frac{-1}{2}$
As we know, Angle of intersection of two curves is given by $\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
When m1 is 4 and m2 is $\frac{-1}{2}$
Then,
\begin{aligned} &\operatorname{Tan} \theta=\left(\frac{4-\left(\frac{-1}{2}\right)}{1+4 \times\left(\frac{-1}{2}\right)}\right) \\ &\operatorname{Tan} \theta=\left(\frac{\left(\frac{9}{2}\right)}{-1}\right) \\ &\tan \theta=\left|\left(\frac{-9}{2}\right)\right| \\ &\tan \theta=\frac{9}{2} \\ &\theta=\tan ^{-1}\left(\frac{9}{2}\right) \end{aligned}
When $m_{1}=-4$ and $m_{2}=\frac{1}{2}$
\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{\left(-4-\frac{1}{2}\right)}{1+(-4)\left(\frac{1}{2}\right)}\right|=\left|\frac{\left(\frac{-8-1}{2}\right)}{1-2}\right|=\left|\frac{\frac{-9}{2}}{-1}\right| \\ &\operatorname{Tan} \theta=\left|\frac{9}{2}\right| \\ &\operatorname{Tan} \theta=\frac{9}{2} \\ &=\theta=\tan ^{-1}\left(\frac{9}{2}\right) \end{aligned}

Tangents and Normals exercise 15.3 , question 1 sub question 3

$\theta=\frac{\pi}{2}$ and $\tan^{-1} \left (\frac{1}{2} \right )$
Hint – The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
m1=slope of first curve. m2=slope of second curve.
Given- $2 y^{2}=x^{3} \ldots \ldots(1)\;\; \& \;\; y^{2}=32 x \ldots \ldots(2)$
First curve is $2y=x^{3}$
Differentiating above with respect to x,
As we know, $\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=4 y \frac{d y}{d x}=3 x^{2} \\ &=m_{1}=\frac{d y}{d x}=\frac{3 x^{2}}{4 y} \ldots \ldots \text { (3) } \end{aligned}
Second curve is $y^{2}=32x$
Differentiating above with respect to x,
\begin{aligned} &=2 y \frac{d y}{d x}=32 \\ &=\frac{d y}{d x}=\frac{32}{2 y}=\frac{16}{y} \\ &=m_{2}=\frac{d y}{d x}=\frac{16}{y} \ldots \ldots \text { (4) } \end{aligned}
Substituting (1) in (2),we get
\begin{aligned} &=2 y^{2}=x^{3} \\ &=2(32 x)=x^{3} \\ &=64 x=x^{3} \\ &=x^{3}-64 x=0 \end{aligned}
\begin{aligned} &=x\left(x^{2}-64\right)=0 \\ &=x=0 \text { or } x^{2}-64=0 \\ &=x=0 \text { or } x^{2}=64 \\ &=x=0 \& x=\pm 8 \end{aligned}
Substituting x=0 or $\pm 8$ in (2)
$y^{2}=32x$ This is not possible
When x=0
\begin{aligned} &y^{2}=32(0) \\ &y=0 \end{aligned}
When x=8
\begin{aligned} &y^{2}=32(8) \\ &y^{2}=256 \\&y=\pm 16 \end{aligned}
Substituting the values for m1& m2 , we get,
When, x=0
\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{3 x^{2}}{4 y} \\ &=m_{1}=\frac{3(0)}{4(16)}=0 \end{aligned}
When x=8,y=16
$=m_{1}=\frac{3(8)^{2}}{4(16)}=3$
Value of m1 is 0 and 3
When x=0 & y=0
\begin{aligned} &=m_{2}=\frac{d y}{d x}=\frac{16}{y} \\ &=m_{2}=\frac{16}{0}=\infty \quad\left\{:: \frac{1}{0}=\infty\right\} \end{aligned}
When y=16
Values of m2 is $\infty$ and 1
As we know, Angle of intersection of two curves is given by $\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
When m1 is 0 and m2 is $\infty$
$\theta=\tan ^{-1}(\infty)$ As we know $\tan \left(\frac{\pi}{2}\right)=\infty$
\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{Tan} \theta=\left(\frac{\infty-0}{1+\infty \times 0}\right) \\ &\operatorname{Tan} \theta=\infty \\ &\theta=\tan ^{-1}(\infty) \\ &=\tan ^{-1}(\infty)=\frac{\pi}{2} \\ &\theta=\frac{\pi}{2} \end{aligned}

$\infty=\frac{\pi}{2}$

When $m_{1}=3$ and $m_{2}=1$
\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{(3-1)}{1+(1)(3)}\right|=\left|\frac{2}{1+3}\right|=\left|\frac{2}{4}\right| \\ &\operatorname{Tan} \theta=\left|\frac{1}{2}\right| \\ &\operatorname{Tan} \theta=\frac{1}{2} \\ &=\theta=\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}

Tangents and Normals exercise 15.3 , question 1 sub question 4

Answer: $\frac{\pi}{4}$
Hint – The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
m1=slope of first curve. m2=slope of second curve.
Given – $x^{2}+y^{2}-4 x-1=0 \ldots \text { (1) }$ &
$x^{2}+y^{2}-2 y-9=0 \ldots(2)$

First curve is $x^{2}+y^{2}-4 x-1=0$
\begin{aligned} &x^{2}-4 x+4+y^{2}-4-1=0 \\ &(x-2)^{2}+y^{2}-5=0 \ldots \ldots \text { (3) } \end{aligned}

Subtracting (2) from (1) we get ,
\begin{aligned} &=x^{2}+y^{2}-4 x-1-\left(x^{2}+y^{2}-2 y-9\right)=0 \\ &=x^{2}+y^{2}-4 x-1-x^{2}-y^{2}+2 y+9=0 \\ &=-4 x-1+2 y+9=0 \\ &=-4 x+2 y+8=0 \\ &=2 y=4 x-8 \\ &=y=2 x-4 \end{aligned}
Substituting y=2x-4 in (3), we get
\begin{aligned} &=(x-2)^{2}+(2 x-4)^{2}-5=0 \\ &=(x-2)^{2}+4(x-2)^{2}-5=0 \\ &=(x-2)^{2}+(1+4)-5=0 \\ &=5(x-2)^{2}-5=0 \end{aligned}
\begin{aligned} &=5\left((x-2)^{2}-1\right)=0 \\ &=(x-2)^{2}-1=0 \\ &=(x-2)^{2}=1 \\ &=(x-2)=\pm 1 \end{aligned}
\begin{aligned} &=x=1+2 \text { or } x=-1+2 \\ &=x=3 \text { or } x=1 \end{aligned}
so When x =3
\begin{aligned} &y=2 \times 3-4 \\ &y=6-4 \\ &y=2 \end{aligned}
when x=1
\begin{aligned} &y=2 \times 1-4 \\ &y=2-4 \\ &y=-2 \end{aligned}
The point of intersection of two curves are (3, 2) & (1,-2).
Differentiating curves (1) & (2) with respect to x,
\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0 \\ &=x^{2}+y^{2}-4 x-1=0 \\ &=2 x+\frac{2 y(d y)}{d x}-4-0=0 \end{aligned}
\begin{aligned} &=x+y \frac{d y}{d x}-2=0 \\ &=y \frac{d y}{d x}=2-x \\ &m_{1}=\frac{d y}{d x}=\frac{2-x}{y} \ldots \ldots \text { (4) } \end{aligned}
Second curve is $x^{2}+y^{2}-2 y-9=0$
\begin{aligned} &2 x+\frac{2 y(d y)}{d x}-\frac{2 d y}{d x}-0=0 \\ &=x+y \frac{d y}{d x}-\frac{d y}{d x}=0 \\ &=x+(y-1) \frac{d y}{d x}=0 \\ &=m_{2}=\frac{d y}{d x}=\frac{-x}{y-1} \ldots \ldots(5) \end{aligned}
At (3, 2) in eq (4), we get
$m_{1}=\frac{d y}{d x}=\frac{2-3}{2}=\frac{-1}{2}$
At (3, 2) in eq (5), we get
$m_{2}=\frac{d y}{d x}=\frac{-3}{2-1}=-3$
At (1,-2) in eq (4), we get
\begin{aligned} &m_{1}=\frac{d y}{d x}=\frac{2-(1)}{-2}=\frac{2}{-2}=-1 \\ &m_{2}=\frac{d y}{d x}=\frac{-x}{y-1}=\frac{-1}{-2-1}=\frac{-1}{-3}=\frac{1}{3} \end{aligned}

Value of $m_{1}=\frac{-1}{2} \&-1$

Value of $m_{2}=-3 \& \frac{1}{3}$
Angle of intersection of two curves is given by $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
When m1is$\frac{-1}{2}$ and m2 is-3
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{tan} \theta=\left|\frac{-\frac{1}{2}-(-1)}{1+\left(-\frac{1}{2}\right)(-1)}\right| \\ &\operatorname{tan} \theta=|-1| \\ &\operatorname{tan} \theta=1 \quad:: \tan \left(\frac{\pi}{4}\right)=1 \\ &=\frac{\pi}{4} \end{aligned}

When m1=-1 and m2=$\frac{1}{3}$
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{-1-\left(\frac{1}{3}\right)}{1+(-1)\left(\frac{1}{3}\right)}\right|=\left|\frac{-1+\frac{1}{3}}{1-\frac{1}{3}}\right| \\ &\operatorname{tan} \theta=|-1| \\ &\operatorname{tan} \theta=1 \\ &=\theta=\frac{\pi}{4} \end{aligned}

Tangents and Normals exercise 15.3 , question 1 sub question 5

$\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)$
Hint - The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
m1=slope of first curve. m2=slope of second curve.
Given –
$\begin{array}{r} \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \ldots \ldots \text { (1) } \\ x^{2}+y^{2}=a b \ldots \ldots \text { (2) } \end{array}$
Considering second curve
\begin{aligned} &x^{2}+y^{2}=a b \\ &y^{2}=a b-x^{2} \end{aligned}
Substituting this in eq (1)
\begin{aligned} &\frac{x^{2}}{a^{2}}+\frac{a b-x^{2}}{b^{2}}=1 \\ &\frac{x^{2} b^{2}+a^{2\left(a b-x^{2}\right)}}{a^{2} b^{2}}=1 \\ &x^{2} b^{2}+a^{3} b-a^{2} x^{2}=a^{2} b^{2} \end{aligned}
\begin{aligned} &x^{2} b^{2}-a^{2} x^{2}=a^{2} b^{2}-a^{3} b \\ &x^{2}\left(b^{2}-a^{2}\right)=a^{2} b(b-a) \\ &x^{2}=\frac{a^{2} b(b-a)}{b^{2}-a^{2}}::\left(x^{2}-y^{2}\right)=(x+y)(x-y) \end{aligned}
\begin{aligned} x^{2} &=\frac{a^{2} b(b-a)}{(b+a)(b-a)} \\ x^{2} &=\frac{a^{2} b}{(b+a)} \\ x &=\pm \sqrt{\frac{a^{2} b}{(b+a)}} \ldots \ldots \text { (3) } \end{aligned}
Since, \begin{aligned} y^{2}=ab-x^{2} \end{aligned}
\begin{aligned} &y^{2}=a b-\frac{a^{2} b}{b+a} \\ &y^{2}=\frac{a b^{2}+a^{2} b-a^{2} b}{b+a} \end{aligned}
\begin{aligned} y^{2} &=\frac{a b^{2}}{b+a} \\ y &=\pm \sqrt{\frac{a b^{2}}{b+a}} \ldots \ldots \text { (4) } \end{aligned}
Since curves are $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \& x^{2}+y^{2}=a b$
Differentiating above with respect to x,
\begin{aligned} &=\frac{2 x^{2}}{a^{2}}+\frac{2 y}{b^{2}} \cdot \frac{d y}{d x}=0 \\ &=\frac{y}{b^{2}} \cdot \frac{d y}{d x}=\frac{-x}{a^{2}} \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{-x}{a^{2}} \times \frac{b^{2}}{y}=\frac{-b^{2} x}{a^{2} y}\\ &=m_{1}=\frac{d y}{d x}=\frac{-b^{2} x}{a^{2} y} \ldots \ldots(5) \end{aligned}
Second curve is \begin{aligned} & x^{2} +y^{2}=ab \end{aligned}
\begin{aligned} &=2 x^{2}+2 y^{2} \cdot \frac{d y}{d x}=0 \\ &=m_{2}=\frac{d y}{d x}=\frac{-x}{y} \ldots \ldots \text { (6) } \end{aligned}
Substituting (3) in (4) above values for m1 & m2, we get
At$\left(\sqrt{\frac{a^{2} b}{b+a}}, \sqrt{\frac{a b^{2}}{b+a}}\right)$ in eq (6), we get
\begin{aligned} &=\frac{d y}{d x}=\frac{-\sqrt{\frac{a^{2} b}{b+a}}}{\sqrt{\frac{a b^{2}}{a+b}}} \\ &=\frac{d y}{d x}=\frac{-a \sqrt{\frac{b}{b+a}}}{b \sqrt{\frac{a}{b+a}}} \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{-a \sqrt{b}}{b \sqrt{a}} \\ &=m_{2}=\frac{d y}{d x}=-\sqrt{\frac{a}{b}} \end{aligned}
When $m_{1}=\frac{-b \sqrt{b}}{a \sqrt{a}}$ and $m_{2}=-\sqrt{\frac{a}{b}}$
Angle of intersection of two curves is given by $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{tan} \theta=\left|\frac{\frac{-b \sqrt{b}}{a \sqrt{a}}--\sqrt{\frac{a}{b}}\left(-\sqrt{\frac{a}{b}}\right)}{1+\left(\frac{-b \sqrt{b}}{a \sqrt{a}}\right)\left(-\sqrt{\frac{a}{b}}\right)}\right| \end{aligned}
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{\frac{-b \sqrt{b} \times \sqrt{b}+a \sqrt{a} \times \sqrt{a}}{a \sqrt{a} \times \sqrt{b}}}{1+\frac{b}{a}}\right| \\ &\operatorname{tan} \theta=\left|\frac{\frac{-b \times b+a \times a}{a \sqrt{a b}}}{1+\frac{b}{a}}\right| \end{aligned}
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{\frac{a^{2}-b^{2}}{a \sqrt{a b}}}{\frac{a+b}{a}}\right| \\ &\operatorname{tan} \theta=\left|\frac{\frac{(a-b)(a+b)}{\sqrt{a b}}}{a+b}\right| \end{aligned}
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{a-b}{\sqrt{a b}}\right| \\ &\operatorname{tan} \theta=\frac{a-b}{\sqrt{a b}} \\ &\theta=\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right) \end{aligned}

Tangents and Normals exercise 15.3 , question 1 sub question 6

$\theta=\tan^{-1}3$
Hint - The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
m1=slope of first curve. m2=slope of second curve.
Given –
\begin{aligned} &x^{2}+4 y^{2}=8 \ldots \ldots(1) \\ &x^{2}-2 y^{2}=2 \ldots \ldots \text { (2) } \end{aligned}
Considering second curve
\begin{aligned} &x^{2}-2 y^{2}=2 \\ &x^{2}=2+2 y^{2} \end{aligned}
Substituting this in eq (1)
\begin{aligned} &2+2 y^{2}+4 y^{2}=8 \\ &6 y^{2}=6 \\ &y^{2}=1 \\ &y=\pm 1 \end{aligned}
Now putting the value of y in $x^{2}=2+2 y^{2}$
When y = 1
\begin{aligned} &x^{2}=2+2(1)^{2} \\ &x^{2}=2+2 \\ &x^{2}=4 \\ &x=\pm 2 \end{aligned}
Point of intersection are (2,1) and (-2,-1)
Since curves are $x^{2}+4 y^{2}=8\;\; \&\;\; x^{2}-2 y^{2}=2$
Differentiating above with respect to x
\begin{aligned} &x^{2}+4 y^{2}=8 \\ &2 x+8 y \frac{d y}{d x}=0 \\ &8 y \frac{d y}{d x}=-2 x \\ &=\frac{d y}{d x}=\frac{-2 x}{8 y} \end{aligned}
\begin{aligned} &m_{1}=\frac{d y}{d x}=\frac{-x}{4 y} \ldots \ldots(3) \\ &\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0 \\ &=x^{2}-2 y^{2}=2 \end{aligned}
\begin{aligned} &=2 x-\frac{4 y(d y)}{d x}=0 \\ &=2 x=4 y \frac{d y}{d x} \\ &=\frac{d y}{d x}=\frac{2 x}{4 y} \\ &m_{2}=\frac{d y}{d x}=\frac{x}{2 y} \ldots \ldots \text { (4) } \end{aligned}
When (2,1)
\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{-x}{4 y}=\frac{-2}{4(1)}=\frac{-2}{4}=\frac{-1}{2} \\ &=m_{1}=\frac{-1}{2} \end{aligned}
When (-2,-1)
\begin{aligned} &=m_{2}=\frac{d y}{d x}=\frac{x}{2 y}=\frac{-2}{2(-1)}=\frac{-2}{-2}=1 \\ &=m_{2}=1 \end{aligned}
Angle of intersection of two curves is given by $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
When m1is $\frac{-1}{2}$ and m2 is 1
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{tan} \theta=\left|\frac{-\frac{1}{2}-1}{1+\left(-\frac{1}{2}\right)(1)}\right| \end{aligned}
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{\frac{-3}{2}}{\frac{1}{2}}\right| \\ &\operatorname{tan} \theta=\left|\frac{-3}{2} \times \frac{2}{1}\right| \\ &\operatorname{tan} \theta=|-3| \\ &\operatorname{tan} \theta=(3) \\ &=\theta=\tan ^{-1} 3 \end{aligned}

Tangents and Normals exercise 15.3 question 1 sub question 7

$\tan ^{-1} \frac{9}{13}$
Hint - The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
m1=slope of first curve. m2=slope of second curve.
Given –
\begin{aligned} &x^{2}=27 y \ldots \ldots(1) \\ &y^{2}=8 x \ldots \ldots(2) \end{aligned}
Considering second curve
\begin{aligned} &y^{2}=8 x \\ &x=\frac{y^{2}}{8} \end{aligned}
Substituting this in eq (1)
\begin{aligned} &\left(\frac{y^{2}}{8}\right)^{2}=27 y \\ &\frac{y^{4}}{64}=27 y \\ &y^{4}=y(27 \times 64) \end{aligned}
\begin{aligned} &y^{4}-y(27 \times 64)=0 \\ &y\left(y^{3}-(27 \times 64)\right)=0 \\ &y=0 \text { or } y^{3}=27 \times 64 \\ &y=0 \text { or } y=3 \times 4 \\ &y=0 \text { or } y=12 \end{aligned}
Now putting the value of y=0 & 12 in $x=\frac{y^{2}}{8}$
When y = 0
$x^{2}=\frac{(0)^{2}}{8}=0$
When y = 12
\begin{aligned} &x^{2}=\frac{12^{2}}{8}=\frac{144}{8}=18 \\ &x=0 \text { or } x=18 \end{aligned}
Point of intersection are (0, 0) and (18, 12)
Since curves are $x^{2}=27 y \;\; \&\;\; y^{2}=8 x$
Differentiating above with respect to x
\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0 \\ &=x^{2}=27 y \\ &=2 x=27 y \frac{d y}{d x} \\ &m_{1}=\frac{d y}{d x}=\frac{2 x}{27 y} \ldots \ldots \text { (3) } \end{aligned}
Now consider $y^{2}=8 x$
\begin{aligned} &2 y \frac{d y}{d x}=\frac{8}{2 y} \\ &=m_{2}=\frac{d y}{d x}=\frac{4}{y} \end{aligned}
When (0,0)
\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{2(0)}{27}=0 \\ &=m_{2}=\frac{d y}{d x}=\frac{4}{0}=\infty \end{aligned}
When (18, 12)
\begin{aligned} &=\frac{2 x}{27}=\frac{2 \times 18}{27}=\frac{4}{3} \\ &=m_{1}=\frac{4}{3} \\ &=m_{2}=\frac{d y}{d x}=\frac{8}{2 v}=\frac{8}{2 \times 12}=\frac{1}{3} \end{aligned}

Value of m1=0 and $\frac{4}{3}$ , m2= $\infty$ and $\frac{1}{3}$

Angle of intersection of two curves is given by $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
When m1 is $\frac{4}{3}$ and m2 is $\frac{1}{3}$
\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{Tan} \theta=\left|\frac{\frac{4}{3}-\frac{1}{3}}{1+\left(\frac{4}{3}\right)\left(\frac{1}{3}\right)}\right| \\ &\operatorname{Tan} \theta=\left|\frac{\frac{3}{4}}{1+\frac{4}{9}}\right| \end{aligned}
\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{3}{3} \times \frac{9}{13}\right| \\ &\operatorname{Tan} \theta=\left|\frac{9}{13}\right| \\ &\operatorname{Tan} \theta=\left(\frac{9}{13}\right) \\ &=\theta=\tan ^{-1} \frac{9}{13} \end{aligned}

Tangents and Normals exercise 15.3 question 1 sub question 8

$tan^{-1}\frac{1}{2}$
Hint - The angle of intersection of curves is $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
m1=slope of first curve. m2=slope of second curve.
Given –
\begin{aligned} &x^{2}+y^{2}=2 x \ldots \ldots(1) \\ &y^{2}=x \ldots \ldots \text { (2) } \end{aligned}
Considering second curve
\begin{aligned} &y^{2}=x \end{aligned}

Substituting this in eq (1)
$x^{2}+x=2x\\ x^{2}+x-2x=0\\ x^{2}-x=0\\ x\left (x-1 \right )=0\\ x = 0\;\;\; or\;\;\; x-1 =0\\ x =0\;\;\; or\;\;\;x = 1\\$

Now putting the value of x = 0 & 1 in \begin{aligned} &y^{2}=x \end{aligned}
When x = 0
$y^{2}=0\\ y=0$
When x = 1
$y^{2}=1\\ y=\pm 1$
Point of intersection are (0, 0) and (1, 1)
Since curves are $x^{2}+y^{2}=2 x \;\;\&\;\; y^{2}=x$
Differentiating above with respect to x
As we know, $\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=x^{2}+y^{2}=2 x \\ &=2 x+2 y \frac{d y}{d x}=2 \\ &=2 y \frac{d y}{d x}=2-2 x \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{2-2 x}{2 y}=\frac{2(1-x)}{2 y} \\ &m_{1}=\frac{d y}{d x}=\frac{1-x}{y} \end{aligned}
Now consider \begin{aligned} &y^{2}=x \end{aligned}
\begin{aligned} &2 y \frac{d y}{d x}=1 \\ &=m_{2}=\frac{d y}{d x}=\frac{1}{2 y} \end{aligned}
When (0, 0)
\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{1-0}{0}=\infty \\ &=m_{2}=\frac{d y}{d x}=\frac{1}{2(0)}=\infty \end{aligned}
When (1, 1)
\begin{aligned} &=m_{1}=\frac{1-1}{1}=0 \\ &=m_{2}=\frac{1}{2(1)}=\frac{1}{2} \end{aligned}

Angle of intersection of two curves is given by$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
When m1 is $\frac{4}{3}$ and m2 is $\frac{1}{3}$
\begin{aligned} &\operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{tan} \theta=\left|\frac{0-\frac{1}{2}}{1+0\left(\frac{1}{2}\right)}\right| \\ &\operatorname{tan} \theta=\left|\frac{\frac{-1}{2}}{1}\right| \\ &\operatorname{tan} \theta=\left(\frac{1}{2}\right) \end{aligned}

Tangents and Normals exercise 15.3 question 1 sub question 9

$\tan^{-1}\frac{4\sqrt{2}}{7}$
Hint - The angle of intersection of curves is$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
m1=slope of first curve. m2=slope of second curve.
Given –
\begin{aligned} &y=4-x^{2} \ldots \ldots(1) \\ &y=x^{2} \ldots \ldots(2) \end{aligned}
Substituting eq (2) in (1) we get
,\begin{aligned} &x^{2}=4-x^{2} \\ &2 x^{2}=4 \\ &x^{2}=2 \\ &x=\pm \sqrt{2} \end{aligned}
Put \begin{aligned} &x=\pm \sqrt{2} \end{aligned} in eq (2), we get
When \begin{aligned} &x=\sqrt{2} \end{aligned}
\begin{aligned} &y=(\sqrt{2})^{2}=2 \\ &y=2 \end{aligned}
When \begin{aligned} &x=-\sqrt{2} \end{aligned}
\begin{aligned} &y=(-\sqrt{2})^{2}=2 \\ &y=2 \end{aligned}
Thus two curves intersect at \begin{aligned} &\left (\sqrt{2},2 \right ) \end{aligned}and \begin{aligned} &\left (-\sqrt{2},2 \right ) \end{aligned}
Since curves are $y=4-x^{2}$ and $y=x^{2}$
Differentiating above with respect to x
As we know, $\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
Consider first curve
\begin{aligned} &=y=4-x^{2} \\ &=\frac{d y}{d x}=0-2 x \\ &m_{1}=\frac{d y}{d x}=-2 x \end{aligned}

Now consider second curve $y=x^{2}$
$=m_{2}=\frac{d y}{d x}=2 x$

When \begin{aligned} &\left (\sqrt{2},2 \right ) \end{aligned}
\begin{aligned} &=m_{1}=\frac{d y}{d x}=-2 x=-2(\sqrt{2})=-2 \sqrt{2} \\ &=m_{1}=-2 \sqrt{2} \end{aligned}

When \begin{aligned} &\left (-\sqrt{2},2 \right ) \end{aligned}
\begin{aligned} &=m_{1}=\frac{d y}{d x}=2 x=2(-\sqrt{2})=-2 \sqrt{2} \\ \end{aligned}
Angle of intersection of two curves is given by $\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
\begin{aligned} &\operatorname{Tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ &\operatorname{Tan} \theta=\left|\frac{2 \sqrt{2}-(-2 \sqrt{2})}{1+(2 \sqrt{2})(-2 \sqrt{2})}\right| \\ &\operatorname{Tan} \theta=\left|\frac{2 \sqrt{2}+2 \sqrt{2}}{1-(4)(2)}\right|=\left|\frac{4 \sqrt{2}}{1-8}\right|=\left|\frac{4 \sqrt{2}}{7}\right| \\ &=\theta=\tan ^{-1} \frac{4 \sqrt{2}}{7} \end{aligned}

Tangents and Normals exercise 15.3 question 2 sub question 1

$m_{1} \times m_{2}=-1$ Hence, two curves intersect orthogonally.
Hint - Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given –
\begin{aligned} &y=x^{3}---(1) \\ &6 y=7-x^{2}---(2) \end{aligned}
Substituting y= x3 in eq (2), we get
\begin{aligned} &=6\left(x^{3}\right)=7-x^{2} \\ &=6 x^{3}+x^{2}-7=0 \end{aligned}
Since \begin{aligned} &f(x)=6x^{3}+x^{2}-7=0 \\ \end{aligned}, we have to find f(x)=0, so that x is a factor of f(x).
When x = 1
\begin{aligned} f(1) &=6(1)^{3}+(1)^{2}-7 \\ &=6+1-7 \\ &=0 \end{aligned}
Hence, x = 1 is a factor of f(x)
Substituting x = 1 in y= x3 , we get
\begin{aligned} &y=(1)^{3} \\ &y=1 \end{aligned}
The point of intersection of two curves is (1, 1)
First curve is \begin{aligned} &y=x^{3} \\ \end{aligned}
Differentiating above with respect to x,
As we know,$\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
$=m_{1}=\frac{d y}{d x}=3 x^{2}$
Second curve is $6y=7-x^{2}$
Differentiating above with respect to x,
\begin{aligned} &=\frac{6 d y}{d x}=0-2 x \\ &=\frac{d y}{d x}=\frac{-2 x}{6} \\ &=m_{2}=\frac{d y}{d x}=\frac{-x}{3} \end{aligned}
Now, put (1,1) in m1 & m2 , we get,
\begin{aligned} &=m_{1}=\frac{d y}{d x}=3 x^{2}=3(1)^{2}=3 \\ &=m_{1}=3 \\ &=m_{2}=\frac{d y}{d x}=\frac{-x}{3}=\frac{-1}{3} \\ &=m_{2}=\frac{-1}{3} \end{aligned}
When m1=3 and m2=\begin{aligned} \frac{-1}{3} \end{aligned}
Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$
$=3\times \frac{-1}{3}=-1$
Hence, two curves $y=x^{3}$ & $6y=7-x^{2}$ intersect orthogonally.

Tangents and Normals exercise 15.3 question 2 sub question 2

$m_{1} \times m_{2}=-1$
Hence, two curves intersect orthogonally.
Hint - Two curves intersect orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given – \begin{aligned} x^{3}-3 x y^{2}=-2---(1) \\ \end{aligned}
$3 x^{2} y-y^{3}=2---(2)$
Adding (1) & (2), we get
\begin{aligned} &=x^{3}-3 x y^{2}+3 x^{2} y-y^{3}=-2+2 \\ &=x^{3}-3 x y^{2}+3 x^{2} y-y^{3}=0 \\ &=(x-y)^{3}=0 \\ &=(x-y)=0 \\ &=x=y \end{aligned}
Substituting x=y in eq(1), we get
\begin{aligned} &=x^{3}-3 x(x)^{2}=-2 \\ &=x^{3}-3 x^{3}=-2 \\ &=-2 x^{3}=-2 \\ &=x^{3}=1 \\ &=x=1 \end{aligned}

Since x=y (y=1)

The point of intersection of two curves is (1, 1)
First curve is $x^{3}-3 x y^{2}=-2$
Differentiating above with respect to x,
As we know, $\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=3 x^{2}-3\left(1 \times y^{2}-x \times 2 y \frac{d y}{d x}\right)=0 \\ &=3 x^{2}-3 y^{2}-6 x y \frac{d y}{d x}=0 \end{aligned}
\begin{aligned} &=3 x^{2}-3 y^{2}=6 x y \frac{d y}{d x} \\ &=\frac{d y}{d x}=\frac{3\left(x^{2}-y^{2}\right)}{6 x y} \\ &=m_{1}=\frac{d y}{d x}=\frac{\left(x^{2}-y^{2}\right)}{2 x y}---(3) \end{aligned}
Second curve is $3 x^{2} y-y^{3}=2$
Differentiating above with respect to x,
\begin{aligned} &=3\left(2 x y+\frac{x^{2} d y}{d x}\right)-3 y^{2} \frac{d y}{d x}=0 \\ &=6 x y+3 x^{2} \frac{d y}{d x}-3 y^{2} \frac{d y}{d x}=0 \end{aligned}
\begin{aligned} &=6 x y+\left(3 x^{2}-3 y^{2}\right) \frac{d y}{d x}=0 \\ &=\frac{d y}{d x}=\frac{-6 x y}{3 x^{2}-3 y^{2}}=\frac{-6 x y}{3\left(x^{2}-y^{2}\right)} \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{-2 x y}{x^{2}-y^{2}} \\ &=m_{2}=\frac{d y}{d x}=\frac{-2 x y}{x^{2}-y^{2}}---(4) \end{aligned}
When $m_{1}=\frac{\left(x^{2}-y^{2}\right)}{2 x y}$ and $m_{2}=\frac{-2 x y}{x^{2}-y^{2}}$
Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$
$=\frac{\left(x^{2}-y^{2}\right)}{2 x y} \times \frac{-2 x y}{x^{2}-y^{2}}=-1$
Two curves $x^{3}-3 x y^{2}=-2 \;\; \& \;\; 3 x^{2} y-y^{3}=2$ intersect orthogonally.

Tangents and Normals exercise 15.3 question 2 sub question 3

$m_{1} \times m_{2}=-1$
Hence, two curves intersect orthogonally.
Hint - Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given – $x^{2}+4y^{2}=8---(1)$
$x^{2}-2y^{2}=4---(2)$
Solving (1) & (2),we get
From (2) curve,
$x^{2}=4+2y^{2}---(1)$
Substituting in (1)
\begin{aligned} &=x^{2}+4 y^{2}=8 \\ &=4+2 y^{2}+4 y^{2}=8 \\ &=6 y^{2}=4 \end{aligned}
\begin{aligned} &=y^{2}=\frac{4}{6} \\ &=y=\pm \sqrt{\frac{2}{3}} \end{aligned}
Substituting $y=\pm \sqrt{\frac{2}{3}}$ in $x^{2}=4+2 y^{2}$ , we get
\begin{aligned} &=x^{2}=4+2\left(\pm \sqrt{\frac{2}{3}}\right)^{2} \\ &=x^{2}=4+2\left(\frac{2}{3}\right) \\ &=x^{2}=4+\frac{4}{3}=\frac{12+4}{3}=\frac{16}{3} \\ &=x=\pm \frac{4}{\sqrt{3}} \end{aligned}

The point of intersection of two curves is $\left(\frac{4}{\sqrt{3}}, \frac{2}{(\sqrt{3})}\right)\;\; \&\;\;\left(-\frac{4}{\sqrt{3}},-\frac{2}{(\sqrt{3})}\right)$

First curve is $x^{2}+4 y^{2}=8$
Differentiating above with respect to x,
As we know, $\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=2 x^{2}+8 y\left(\frac{d y}{d x}\right)=0 \\ &=8 y \frac{d y}{d x}=-2 x \\ &=\frac{d y}{d x}=\frac{-2 x}{8 y}=\frac{-x}{4 y} \\ &=m_{1}=\frac{d y}{d x}=\frac{-x}{4 y}---(3) \end{aligned}
Second curve is $x^{2}-2 y^{2}=4$
Differentiating above with respect to x,
\begin{aligned} &=2 x-4 y \frac{d y}{d x}=0 \\ &=4 y \frac{d y}{d x}=2 x \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{2 x}{4 y}=\frac{x}{2 y} \\ &=m_{2}=\frac{d y}{d x}=\frac{x}{2 y}---(4) \end{aligned}
At $\left(\frac{4}{\sqrt{3}}, \frac{2}{(\sqrt{3})}\right)$ in eq (3), we get
\begin{aligned} &=\frac{d y}{d x}=\frac{\frac{-4}{\sqrt{3}}}{4 \times\left(\sqrt{\frac{2}{3}}\right)} \\ &=\frac{d y}{d x}=\frac{\frac{-1}{\sqrt{3}}}{\left(\sqrt{\frac{2}{3}}\right)} \end{aligned}
\begin{aligned} &=m_{1}=\frac{-1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{2}}=\left(\frac{-1}{\sqrt{2}}\right) \\ &=m_{1}=\frac{-1}{\sqrt{2}} \end{aligned}
At $\left(\frac{4}{\sqrt{3}}, \frac{2}{(\sqrt{3})}\right)$ in eq (4), we get
\begin{aligned} &=\frac{d y}{d x}=\frac{\frac{4}{\sqrt{3}}}{2 \times\left(\sqrt{\frac{2}{3}}\right)} \\ &=m_{2}=\frac{4}{\sqrt{3}} \times \frac{\sqrt{3}}{2 \sqrt{2}}=\left(\frac{2}{\sqrt{2}}\right)=\sqrt{2} \\ &=m_{2}=\sqrt{2} \end{aligned}

When $m_{1}=\frac{-1}{\sqrt{2}}$ and $m_{2}=\sqrt{2}$

Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$
$=\frac{-1}{\sqrt{2}} \times \sqrt{2}=-1$
Two curves $x^{2}+4 y^{2}=8 \;\;\; \&\;\;\; x^{2}-2 y^{2}=4$ intersect orthogonally.

Tangents and Normals exercise 15.3 question 3 sub question 1

$m_{1} \times m_{2}=-1$
Hence, two
curves intersect orthogonally.
Hint - Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given – $x^{2}=4y-------(1)$
$4y+x^{2}-------(2)$
The point of intersection of two curve (2,1).
First curve is $x^{2}=4y$
Differentiating above with respect to x,
As we know,$\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=2 x=4 \frac{d y}{d x} \\ &=\frac{d y}{d x}=\frac{2 x}{4}=\frac{x}{2} \\ &=m_{1}=\frac{d y}{d x}=\frac{x}{2}---(3) \end{aligned}
Second curve is $4y+x^{2}=8$
Differentiating above with respect to x,
\begin{aligned} &=\frac{4 d y}{d x}+2 x=0 \\ &=\frac{d y}{d x}=\frac{-2 x}{4}=\frac{-x}{2} \\ &=m_{2}=\frac{d y}{d x}=\frac{-x}{2}---(4) \end{aligned}
Substituting (2, 1) for m1& m2 , we get,
\begin{aligned} &=m_{1}=\frac{x}{2}=\frac{2}{2}=1 \\ &=m_{1}=1 \\ &=m_{2}=\frac{-x}{2}=\frac{-2}{2}=-1 \\ &=m_{2}=-1 \end{aligned}
When m1=1 and m2=-1
Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$
$= 1 \times -1 =-1$
Hence, two curves $x^{2}=4 y \;\;\;\& \;\;\; 4 y+x^{2}=8$ intersect orthogonally.

Tangents and Normals exercise 15.3 question 3 sub question 2

$m_{1} \times m_{2}=-1$
Hence, two
curves intersect orthogonally.
Hint - Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given – $x^{2}=y-------(1)$
$x^{3}+6y=7-------(2)$
The point of intersection of two curve (1,1).
First curve is $x^{2}=y$
Differentiating above with respect to x,
As we know,$\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=2 x=\frac{d y}{d x} \\ &=m_{1}=2 x---(3) \end{aligned}
Second curve is $x^{3}+6y=7$
Differentiating above with respect to x,
\begin{aligned} &=3 x^{2}+\frac{6 d y}{d x}=0 \\ &=\frac{d y}{d x}=\frac{-3 x^{2}}{6}=\frac{-x^{2}}{2} \\ &=m_{2}=\frac{-x^{2}}{2}---(4) \end{aligned}
Substituting (1, 1) for m1& m2 , we get,
\begin{aligned} &=m_{1}=2 x=2(1)=2 \\ &=m_{1}=2 \\ &=m_{2}=\frac{-x^{2}}{2}=\frac{(-1)^{2}}{2}=\frac{-1}{2} \\ &=m_{2}=\frac{-1}{2} \end{aligned}
When m1=2 and m2=\begin{aligned}\frac{-1}{2} \end{aligned}
Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$
\begin{aligned}=2 \times \frac{-1}{2}=-1 \end{aligned}
Hence, two curves $x^{2}= y \;\;\;\& \;\;\; x^{3}+6y=7$ intersect orthogonally.

Tangents and Normals exercise 15.3 question 3 sub question 3

$m_{1} \times m_{2}=-1$
Hence, two
curves intersect orthogonally.
Hint - Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given – $y^{2}=8x-------(1)$
$2x^{2}+y^{2}=10-------(2)$
The point of intersection of two curve $(1,2\sqrt{2})$.
First curve is $y^{2}=8x$
Differentiating above with respect to x,
As we know,$\frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=2 y \frac{d y}{d x}=8 \\ &=\frac{d y}{d x}=\frac{8}{2 y} \\ &=m_{1}=\frac{4}{y}---(3) \end{aligned}
Second curve is $2x^{2}+y^{2}=10$
Differentiating above with respect to x,
\begin{aligned} &=4 x+\frac{2 y d y}{d x}=0 \\ &=2 y \frac{d y}{d x}=-4 x \\ &=\frac{d y}{d x}=\frac{-4 x}{2 y}=\frac{-2 x}{y} \\ &=m_{2}=\frac{-2 x}{y}---(4) \end{aligned}
Substituting $(1,2\sqrt{2})$ for m1& m2 , we get,
\begin{aligned} &=m_{1}=\frac{4}{y}=\frac{4}{2 \sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2} \\ &=m_{1}=\sqrt{2} \\ &=m_{2}=\frac{-2 x}{y}=\frac{-2 \times 1}{2 \sqrt{2}}=\frac{-1}{\sqrt{2}} \\ &=m_{2}=\frac{-1}{\sqrt{2}} \end{aligned}
When $m_{1}=\sqrt{2}$ and m_{2}=\begin{aligned}\frac{-1}{\sqrt{2}} \end{aligned}
Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$
\begin{aligned}=\sqrt{2} \times \frac{-1}{\sqrt{2}}=-1 \end{aligned}
Hence, two curves $y^{2}=8x \;\;\;\& \;\;\; 2x^{2}+y^{2}=10$ intersect orthogonally.

Tangents and Normals exercise 15.3 question 4

Hence, prove two curves $4x=y^{2}$ and $4xy=k$ cut at right angle ,if $k^{2}=512$.
Hint - Two curves intersect orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given – $4x=y^{2}---(1)$
$4xy=k---(2)$
Prove -
Two curves cut at right angle ,if $k^{2}=512$
Consider First curve is $4x=y^{2}$
Differentiating above with respect to x,
As we know, $\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=4=2 y \frac{d y}{d x} \\ &=\frac{d y}{d x}=\frac{4}{2 y}=\frac{2}{y} \\ &=m_{1}=\frac{d y}{d x}=\frac{2}{y}---(3) \end{aligned}
Second curve is $4xy=k$
Differentiating above with respect to x,
\begin{aligned} &=4\left(y+x \frac{d y}{d x}\right)=0 \\ &=\left(y+x \frac{d y}{d x}\right)=0 \\ &=\frac{x d y}{d x}=-y=\frac{-y}{x} \\ &=m_{2}=\frac{d y}{d x}=\frac{-y}{x}---(4) \end{aligned}
Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$
Since m1& m2 cuts orthogonally
\begin{aligned} &=\frac{2}{y} \times \frac{-y}{x}=-1 \\ &=\frac{-2}{x}=-1 \\ &=x=2 \end{aligned}
Now solving (1) & (2), we get
$= 4xy = k$ & $4x=y^{2}$
Substituting $4x=y^{2}$ in $4xy=k$ ,we get
\begin{aligned} &=\left(y^{2}\right) y=k \\ &=y^{3}=k \\ &=y=k^{\frac{1}{3}} \quad\left\{:: a^{m}=n, a=n^{\frac{1}{m}}\right\} \end{aligned}
Substituting $y=k^{\frac{1}{3}}$ in $4x=y^{2}$ we get
$=4x=\left (k^{\frac{1}{3}} \right )$ and put x=2
\begin{aligned} &=4 \times 2=k^{\frac{2}{3}} \\ &=8=k^{\frac{2}{3}} \end{aligned}
\begin{aligned} &=k^{\frac{2}{3}}=8 \end{aligned} , cube both sides
\begin{aligned} &=\left (k^{\frac{2}{3}} \right )^{3}=8^{3}\\ &=k^{2}=512 \end{aligned}

Tangents and Normals exercise 15.3 question 5

Hence, prove two curves $2x=y^{2}$ and $2xy=k$ cut at right angle ,if $k^{2}=8$.
Hint - Two curves intersect orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given – $2x=y^{2}---(1)$
$2xy=k---(2)$
Prove -
Two curves cut at right angle ,if $k^{2}=8$
Consider First curve is $2x=y^{2}$
Differentiating above with respect to x,
As we know, $\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=2=2 y \frac{d y}{d x} \\ &=2 y \frac{d y}{d x}=2 \\ &=\frac{d y}{d x}=\frac{2}{2 y}=\frac{1}{y} \\ &=m_{1}=\frac{d y}{d x}=\frac{1}{y}---(3) \end{aligned}
Second curve is $2xy=k$
Differentiating above with respect to x,
\begin{aligned} &=2\left(y+x \frac{d y}{d x}\right)=0 \\ &=\left(y+x \frac{d y}{d x}\right)=0 \\ &=\frac{d y}{d x}=\frac{-y}{x} \\ &=m_{2}=\frac{d y}{d x}=\frac{-y}{x}---(4) \end{aligned}
Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$
Since m1& m2 cuts orthogonally
\begin{aligned} &=\frac{1}{y} \times \frac{-y}{x}=-1 \\ &=\frac{-1}{x}=-1 \\ &=x=1 \end{aligned}
Now solving (1) & (2), we get
$= 2x=y^{2}$ & $2xy = k$
Substituting $2x=y^{2}$ in $2xy = k$ ,we get
\begin{aligned} &=\left(y^{2}\right) y=k \\ &=y^{3}=k \\ &=y=k^{\frac{1}{3}} \quad\left\{:: a^{m}=n, a=n^{\frac{1}{m}}\right\} \end{aligned}
Substituting $y=k^{\frac{1}{3}}$ in $2x=y^{2}$ and x=1 we get
$=2x=\left (k^{\frac{1}{3}} \right )^{2}$
\begin{aligned} &=2 \times 1=k^{\frac{2}{3}} \\ &=2=k^{\frac{2}{3}} \end{aligned}
\begin{aligned} &=k^{\frac{2}{3}}=2 \end{aligned} , cube both sides
\begin{aligned} &=\left (k^{\frac{2}{3}} \right )^{3}=2^{3}\\ &=k^{2}=8 \end{aligned}
Hence, prove two curves cut at right angles if \begin{aligned} &k^{2}=8 \end{aligned}

Tangents and Normals exercise 15.3 question 6

Hence, prove two curves $xy=4$ and $x^{2}+y^{2}=8$ touch each other
Hint - Two curves intersect orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given – $xy=4---(1)$
$x^{2}+y^{2}=8---(2)$
Prove -
Consider First curve is $xy=4$
$x=\frac{4}{y}$
Substituting $x=\frac{4}{y}$ in eq (2), we get
\begin{aligned} &=\left(\frac{4}{y}\right)^{2}+y^{2}=8 \\ &=\frac{16}{y^{2}}+y^{2}=8 \\ &=\frac{16+y^{2}\left(y^{2}\right)}{y^{2}}=8 \end{aligned}
\begin{aligned} &=16+y^{4}=8 y^{2} \\ &=y^{4}-8 y^{2}+16=0 \\ &=\left(y^{2}-4\right)^{2}=0 \quad\left\{::(a-b)^{2}=a^{2}+b^{2}+2 a b\right\} \\ &=y^{2}=4 \\ &=y=\pm 2 \end{aligned}
Now substituting value of \begin{aligned}y=\pm 2 \end{aligned} in eq \begin{aligned} x=\frac{y}{4} \end{aligned}
When y = 2
\begin{aligned} x=\frac{4}{2}=2 \end{aligned}
When y = -2
\begin{aligned} x=\frac{4}{-2}=-2 \end{aligned}
Thus, two curves intersect at (2, 2) and (-2,-2)
Consider first curve xy = 4
Differentiating above with respect to y,
As we know, $\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=y+x \frac{d y}{d x}=0 \\ &=x \frac{d y}{d x}=-y \\ &=\frac{d y}{d x}=\frac{-y}{x} \\ &=m_{1}=\frac{d y}{d x}=\frac{-y}{x}---(3) \end{aligned}
Second curve is \begin{aligned} &x^{2}+y^{2}=8\end{aligned}
Differentiating above with respect to y,
\begin{aligned} &=2 x+2 y \frac{d y}{d x}=0 \\ &=\left(2 y \frac{d y}{d x}\right)=-2 x \\ &=\frac{d y}{d x}=\frac{-2 x}{2 y}=\frac{-x}{y} \\ &=m_{2}=\frac{d y}{d x}=\frac{-x}{y}---(4) \end{aligned}
At (2,2) in eq (3) , we get
\begin{aligned} &=m_{1}=\frac{-2}{2}=-1 \\ &=m_{1}=-1 \end{aligned}
At (2, 2) in eq (4) , we get
\begin{aligned} &=m_{1}=\frac{-2}{2}=-1 \\ &=m_{1}=-1 \end{aligned}
Clearly, $\left(\frac{d y}{d x}\right)_{c_{1}}=\left(\frac{d y}{d x}\right)_{c_{2}}$ at $(2,2)$
At (-2,-2) in eq (3) , we get
\begin{aligned} &=m_{1}=\frac{-(-2)}{-2}=-1 \\ &=m_{1}=-1 \end{aligned}
At (-2,-2) in eq (4) , we get
\begin{aligned} &=m_{1}=\frac{-(-2)}{-2}=-1 \\ &=m_{1}=-1 \end{aligned}
Clearly, $\left(\frac{d y}{d x}\right)_{c_{1}}=\left(\frac{d y}{d x}\right)_{c_{2}}$ at $(-2,-2)$
So, given two curves touch each other at (2,2).
Simillarly, it can be seen that two curves touch each other at (-2, -2)

Tangents and Normals exercise 15.3 question 8 sub question 1

$b^{2}=a^{2}$
Hint –
Two curves intersects orthogonally if $m_{1}\times m_{2}=-1$, where $m_{1}$ and $m_{2}$ are the slopes of two curves.
Given –
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1---(1) \\ x y=c^{2}---(2)$
Consider First curve is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1---(1) \\$
Differentiating above with respect to x,
As we know,$\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=\frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \frac{d y}{d x}=0 \\ &=\frac{2 y}{b^{2}} \frac{d y}{d x}=\frac{2 x}{a^{2}} \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{2 x}{a^{2}} \times \frac{b^{2}}{2 y} \\ &=\frac{d y}{d x}=\frac{b^{2} x}{a^{2} y} \\ &=m_{1}=\frac{d y}{d x}=\frac{b^{2} x}{a^{2} y}---(3) \end{aligned}
Second curve is $xy=c^{2}$
Differentiating above with respect to x,
\begin{aligned} &=x \frac{d y}{d x}+y=0 \\ &=\frac{d y}{d x}=\frac{-y}{x} \\ &=m_{2}=\frac{d y}{d x}=\frac{-y}{x}---(4) \end{aligned}
Two curves intersects orthogonally if $m_{1}$ & $m_{2}$ =-1
Since $m_{1}$ & $m_{2}$ cuts orthogonally
\begin{aligned} &=\frac{b^{2} x}{a^{2} y} \times \frac{-y}{x}=-1 \\ &=b^{2}=a^{2} \end{aligned}

Tangents and Normals exercise 15.3 question 8 subquestion 2

$a^{2}-b^{2}=A^{2}+B^{2}$Hint –
Two curves intersects orthogonally if $m_{1}m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given –
\begin{aligned} &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1---(1) \\ &\quad \frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1---(2) \end{aligned}
Consider First curve \begin{aligned} &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \end{aligned}
Differentiating above with respect to x,
As we know,$\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=\frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \frac{d y}{d x}=0 \\ &=\frac{2 y}{b^{2}} \frac{d y}{d x}=\frac{-2 x}{a^{2}} \\ \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{-2 x}{a^{2}} \times \frac{b^{2}}{2 y} \\ &=\frac{d y}{d x}=\frac{-b^{2} x}{a^{2} y} \\ &=m_{1}=\frac{d y}{d x}=\frac{-b^{2} x}{a^{2} y}--(3)\end{aligned}
Consider second curve \begin{aligned} \frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1 \end{aligned}
Differentiating above with respect to x,
\begin{aligned} &=\frac{2 x}{A^{2}}-\frac{2 y}{B^{2}} \frac{d y}{d x}=0 \\ &=\frac{2 y}{B^{2}} \frac{d y}{d x}=\frac{2 x}{A^{2}} \\ \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{2 x}{A^{2}} \times \frac{B^{2}}{2 y} \\ &=\frac{d y}{d x}=\frac{B^{2} x}{A^{2} y} \\ &=m_{2}=\frac{d y}{d x}=\frac{B^{2} x}{A^{2} y}---(4) \end{aligned}
Two curves intersects orthogonally if $m_{1}\times m_{2}=-1$
\begin{aligned} &=\frac{-b^{2} x}{a^{2} y} \times \frac{B^{2} x}{A^{2} y}=-1\\ &=\frac{x^{2} b^{2} B^{2}}{y^{2} a^{2} A^{2}}=1\\ &=\frac{x^{2}}{y^{2}}=\frac{\left(a^{2} A^{2}\right)}{\left(b^{2} B^{2}\right)}---(5) \end{aligned}

Now, subtract eq (1) & (2), we get

\begin{aligned} &=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1-1 \\ &=x^{2}\left(\frac{1}{a^{2}}-\frac{1}{A^{2}}\right)+y^{2}\left(\frac{1}{b^{2}}+\frac{1}{B^{2}}\right)=0 \\ \end{aligned}
\begin{aligned} &=x^{2}\left(\frac{A^{2}-a^{2}}{a^{2} A^{2}}\right)+y^{2}\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right)=0 \\ &=x^{2}\left(\frac{A^{2}-a^{2}}{a^{2} A^{2}}\right)=-y^{2}\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right) \end{aligned}
\begin{aligned} &=\frac{x^{2}}{y^{2}}=-\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right) \times\left(\frac{a^{2} A^{2}}{A^{2}-a^{2}}\right) \\ &=\frac{x^{2}}{y^{2}}=-\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right) \times(-1)\left(\frac{a^{2} A^{2}}{a^{2}-A^{2}}\right) \\ &=\frac{x^{2}}{y^{2}}=-\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right) \times\left(\frac{a^{2} A^{2}}{a^{2}-A^{2}}\right)---(6) \end{aligned}
Put value of $\frac{x^{2}}{y^{2}}$ in eq (5), we get
\begin{aligned} &=\left(\frac{a^{2} A^{2}}{a^{2}-A^{2}}\right)\left(\frac{B^{2}+b^{2}}{b^{2} B^{2}}\right)=\frac{\left(a^{2} A^{2}\right)}{\left(b^{2} B^{2}\right)} \\ &=\left(\frac{B^{2}+b^{2}}{a^{2}-A^{2}}\right)=1 \\ &=B^{2}+b^{2}=a^{2}-A^{2} \\ &=a^{2}-b^{2}=A^{2}+B^{2} \end{aligned}

Tangents and Normals exercise 15.3 question 9

$m_{1} \times m_{2}=-1$ hence, the two curves intersect at right angles.
Hint –
Two curves intersects orthogonally if $m_{1} \times m_{2}=-1$, where m1 and m2 are the slopes of two curves.
Given –
$\begin{gathered} \frac{x^{2}}{a^{2}+d_{1}}+\frac{y^{2}}{b^{2}+d_{1}}=1---(1) \\ \quad \frac{x^{2}}{a^{2}+d_{2}}+\frac{y^{2}}{b^{2}+d_{2}}=1---(2) \end{gathered}$
Consider First curve is $\begin{gathered} \frac{x^{2}}{a^{2}+d_{1}}+\frac{y^{2}}{b^{2}+d_{1}}=1 \\ \end{gathered}$
Differentiating above with respect to x,
As we know, $\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=\frac{2 x}{a^{2}+d_{1}}+\frac{2 y}{b^{2}+d_{1}} \frac{d y}{d x}=0 \\ &=\frac{2 y}{b^{2}+d_{1}} \frac{d y}{d x}=\frac{-2 x}{a^{2}+d_{1}} \\ \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{-x\left(b^{2}+d_{1}\right)}{y\left(a^{2}+d_{1}\right)}\\ &=m_{1}=\frac{d y}{d x}=\frac{-x\left(b^{2}+d_{1}\right)}{y\left(a^{2}+d_{1}\right)}---(3) \end{aligned}
Second curve is $\begin{gathered}\quad \frac{x^{2}}{a^{2}+d_{2}}+\frac{y^{2}}{b^{2}+d_{2}}=1 \end{gathered}$
Differentiating above with respect to x,
\begin{aligned} &=\frac{2 x}{a^{2}+d_{2}}+\frac{2 y}{b^{2}+d_{2}} \frac{d y}{d x}=0 \\ &=\frac{2 y}{b^{2}+d_{2}} \frac{d y}{d x}=\frac{-2 x}{a^{2}+d_{2}} \\ \end{aligned}
\begin{aligned} &=\frac{d y}{d x}=\frac{-x\left(b^{2}+d_{2}\right)}{y\left(a^{2}+d_{2}\right)} \\ &=m_{2}=\frac{d y}{d x}=\frac{-x\left(b^{2}+d_{2}\right)}{y\left(a^{2}+d_{2}\right)}---(4) \end{aligned}
Now, subtract eq (2) from (1), we get
\begin{aligned} &\frac{x^{2}}{a^{2}+d_{1}}+\frac{y^{2}}{b^{2}+d_{1}}-\frac{x^{2}}{a^{2}+d_{2}}-\frac{y^{2}}{b^{2}+d_{2}}=1-1 \\ &=x^{2}\left(\frac{1}{a^{2}+d_{1}}-\frac{1}{a^{2}+d_{2}}\right)+y^{2}\left(\frac{1}{b^{2}+d_{1}}-\frac{1}{b^{2}+d_{2}}\right)=0 \\ \end{aligned}
\begin{aligned} &=x^{2}\left(\frac{d_{2}-d_{1}}{\left(a^{2}+d_{1}\right)\left(a^{2}+d_{2}\right)}\right)=-y^{2}\left(\frac{d_{2}-d_{1}}{\left(b^{2}+d_{1}\right)\left(b^{2}+d_{2}\right)}\right) \\ &=\frac{x^{2}}{y^{2}}=-\left\{\left(\frac{d_{2}-d_{1}}{\left(b^{2}+d_{1}\right)\left(b^{2}+d_{2}\right)}\right) \times \frac{\left(a^{2}+d_{1}\right)\left(a^{2}+d_{2}\right)}{d_{2}-d_{1}}\right\} \\ \end{aligned}
\begin{aligned}&=\frac{x^{2}}{y^{2}}=-\left\{\frac{\left(d_{2}-d_{1}\right)\left(a^{2}+d_{1}\right)\left(a^{2}+d_{2}\right)}{\left(b^{2}+d_{1}\right)\left(b^{2}+d_{2}\right)\left(d_{1}-d_{2}\right)}\right\}---(5) \end{aligned}

Two curves intersect orthogonally if $m_{1}m_{2}=-1$

From (3) & (4)
\begin{aligned} &=\frac{-x}{y} \frac{\left(b^{2}+d_{1}\right)}{\left(a^{2}+d_{1}\right)} \times \frac{-x}{y} \frac{\left(b^{2}+d_{2}\right)}{\left(a^{2}+d_{2}\right)} \\ &=\frac{x^{2}}{y^{2}}=\frac{\left(b^{2}+d_{1}\right)}{\left(a^{2}+d_{1}\right)} \frac{\left(b^{2}+d_{2}\right)}{\left(a^{2}+d_{2}\right)}---(6) \end{aligned}
Now putting the value of $\frac{x^{2}}{y^{2}}$ from eq (5) in eq (6)
Then,
\begin{aligned} &=\left\{\frac{\left(d_{2}-d_{1}\right)\left(a^{2}+d_{1}\right)\left(a^{2}+d_{2}\right)}{\left(b^{2}+d_{1}\right)\left(b^{2}+d_{2}\right)\left(d_{1}-d_{2}\right)}\right\}=\frac{\left(b^{2}+d_{1}\right)}{\left(a^{2}+d_{1}\right)} \frac{\left(b^{2}+d_{2}\right)}{\left(a^{2}+d_{2}\right)} \\ \end{aligned}
\begin{aligned} &=\frac{\left(d_{2}-d_{1}\right)}{\left(d_{1}-d_{2}\right)} \\ &=-\frac{\left(d_{1}-d_{2}\right)}{\left(d_{1}-d_{2}\right)} \\ &=-1 \end{aligned}
So, the curves intersect at right angles.

Tangents and Normals exercise 15.3 question 10

Hence Prove,\begin{aligned} &a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=P^{2} \end{aligned}
Hint –
The equation of tangent at $p(x_{1},y_{1})$
$=\frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}=1$
Given –
$x \cos \alpha+y \sin \alpha=\mathrm{p}---(\mathrm{i})$
And the curve $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Or $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}---(i i)$
Let line and curve touches each other at point $p(x_{1},y_{1})$
$=x_{1} \cos \alpha+y_{1} \sin \alpha=\mathrm{p}---(\mathrm{iii})$
And $b^{2} x_{1}^{2}+a^{2} y_{1}^{2}=a^{2} b^{2}---(i v)$
Differentiating (ii), we get
As we know,$\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0$
\begin{aligned} &=2 b^{2} x+2 a^{2} y \frac{d y}{d x}=0 \\ &=\left(\frac{d y}{d x}\right)_{x_{1}, y_{1}}=\frac{-b^{2} x_{1}}{a^{2} y_{1}} \end{aligned}
Also, slope of line (i) is = $\frac{-\cos \alpha}{\sin \alpha}$
According to question,
\begin{aligned} &=\frac{-b^{2} x_{1}}{a^{2} y_{1}}=\frac{-\cos \alpha}{\sin \alpha} \\ &=\frac{x_{1}}{a^{2} \cos \alpha}=\frac{y_{1}}{b^{2} \sin \alpha}=\partial \\ \end{aligned}
\begin{aligned} &=x_{1}=\partial a^{2} \cos ^{2} \alpha, y c g f y \partial b^{2} \sin ^{2} \alpha=p \\ &=\partial=\frac{p}{a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha} \\ \end{aligned}
\begin{aligned} &=x_{1}=\frac{p a^{2} \cos \alpha}{a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha} \end{aligned} and \begin{aligned} y_{1}=\frac{p b^{2} \sin \alpha}{a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha} \end{aligned}
Putting these values in (iv), we get
\begin{aligned} &=\frac{b^{2} p^{2} a^{4} \cos ^{2} \alpha}{\left(\left(a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\right)\right)^{2}}+\frac{a^{2} p^{2} b^{4} \sin ^{2} \alpha}{\left(\left(a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\right)\right)^{2}}=a^{2} b^{2} \\ \end{aligned}\begin{aligned}&=\frac{a^{2} b^{2} p^{2}\left(a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\right)}{\left(a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\right)^{2}}=a^{2} b^{2} \\ &=a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=P^{2} \end{aligned}

The 15th chapter, Tangents, and Normals in class 12 mathematics, have three exercises. The last exercise, 5.3, consists of concepts like the angle of intersection, curves intersecting orthogonally, proving that the curves touch each other, condition of the set of curves, and curves intersecting at right angles. There are around 23 questions along with their subparts given in this exercise. With the help of RD Sharma Class 12 Chapter 15 exercise 15.3 solution book, the students can easily solve the sums.

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## RD Sharma Chapter wise Solutions

1. What is the prescribed solution book for the students to clarify their doubts regarding the RD Sharma Class 12 Solutions Tangents and Normals Ex 15.3?

The students can use the RD Sharma Class 12th exercise 15.3 solution book to clear their doubts regarding the sums in this chapter.

The RD Sharma solution books can be downloaded from the Career 360 website. All the answers are present in an order-wise manner as given in the textbook.

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The RD Sharma solution books contain the sums solved in various methods. Therefore, the students can decide the way that they wish to adapt.

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All the 23 questions given in the textbook in exercise 15.3 are answered in the RD Sharma Class 12th exercise 15.3 book.

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