NCERT Solutions for Exercise 6.3 Class 12 Maths Chapter 10 - Application of Derivatives

Question:2 . Find the slope of the tangent to the curve $\frac{x-1}{x-2},x\ne 2atx=10$

Answer:

Given curve is,

$y=\frac{x-1}{x-2}$
The slope of the tangent at x = 10 is given by
${\left(\frac{dy}{dx}\right)}_{x=10}=\frac{(1)(x-2)-(1)(x-1)}{(x-2{)}^2}=\frac{x-2-x+1}{(x-2{)}^2}=\frac{-1}{(x-2{)}^2}$
at x = 10
$=\frac{-1}{(10-2{)}^2}=\frac{-1}{8^2}=\frac{-1}{64}$
hence, slope of tangent at x = 10 is $\frac{-1}{64}$

Question:3 Find the slope of the tangent to curve $y=x^3-x+1$ at the point whose x-coordinate is 2.

Answer:

Given curve is,
$y=x^3-x+1$
The slope of the tangent at x = 2 is given by
${\left(\frac{dy}{dx}\right)}_{x=2}=3x^2-1=3(2{)}^2-1=3\times 4-1=12-1=11$
Hence, the slope of the tangent at point x = 2 is 11

Question:4 Find the slope of the tangent to the curve $y=x^3-3x+2$ at the point whose x-coordinate is 3.

Answer:

Given curve is,
$y=x^3-3x+2$
The slope of the tangent at x = 3 is given by
${\left(\frac{dy}{dx}\right)}_{x=3}=3x^2-3=3(3{)}^2-3=3\times 9-3=27-3=24$
Hence, the slope of tangent at point x = 3 is 24

Question:5 Find the slope of the normal to the curve $x=a{\cos }^3\theta ,y=a{\sin }^3\theta at\theta =\pi /4$

Answer:

The slope of the tangent at a point on a given curve is given by
$\left(\frac{dy}{dx}\right)$
Now,
${\left(\frac{dx}{d\theta }\right)}_{\theta =\frac{\pi }{4}}=3a{\cos }^2\theta (-\sin \theta )=3a(\frac{1}{\sqrt{2}}{)}^2(-\frac{1}{\sqrt{2}})=-\frac{3\sqrt{2}a}{4}$
Similarly,
${\left(\frac{dy}{d\theta }\right)}_{\theta =\frac{\pi }{4}}=3a{\sin }^2\theta (\cos \theta )=3a(\frac{1}{\sqrt{2}}{)}^2(\frac{1}{\sqrt{2}})=\frac{3\sqrt{2}a}{4}$
$\left(\frac{dy}{dx}\right)=\frac{\left(\frac{dy}{d\theta }\right)}{\left(\frac{dx}{d\theta }\right)}=\frac{\frac{3\sqrt{2}a}{4}}{-\frac{3\sqrt{2}a}{4}}=-1$
Hence, the slope of the tangent at $\theta =\frac{\pi }{4}$ is -1
Now,
Slope of normal = $-\frac{1}{slopeoftangent}$ = $-\frac{1}{-1}=1$
Hence, the slope of normal at $\theta =\frac{\pi }{4}$ is 1

Question:6 Find the slope of the normal to the curve $x=1-a\sin \theta ,y=b{\cos }^2\theta at\theta =\pi /2$

Answer:

The slope of the tangent at a point on given curves is given by
$\left(\frac{dy}{dx}\right)$
Now,
${\left(\frac{dx}{d\theta }\right)}_{\theta =\frac{\pi }{2}}=-a(\cos \theta )$
Similarly,
${\left(\frac{dy}{d\theta }\right)}_{\theta =\frac{\pi }{2}}=2b\cos \theta (-\sin \theta )$
${\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{2}}=\frac{\left(\frac{dy}{d\theta }\right)}{\left(\frac{dx}{d\theta }\right)}=\frac{-2b\cos \theta \sin \theta }{-a\cos \theta }=\frac{2b\sin \theta }{a}=\frac{2b\times 1}{a}=\frac{2b}{a}$
Hence, the slope of the tangent at $\theta =\frac{\pi }{2}$ is $\frac{2b}{a}$
Now,
Slope of normal = $-\frac{1}{slopeoftangent}$ = $-\frac{1}{\frac{2b}{a}}=-\frac{a}{2b}$
Hence, the slope of normal at $\theta =\frac{\pi }{2}$ is $-\frac{a}{2b}$

Question:7 Find points at which the tangent to the curve $y=x^3-3x^2-9x+7$ is parallel to the x-axis.

Answer:

We are given :

$y=x^3-3x^2-9x+7$

Differentiating the equation with respect to x, we get :

$\frac{dy}{dx}=3x^2-6x-9+0$

or $=3(x^2-2x-3)$

or $\frac{dy}{dx}=3(x+1)(x-3)$

It is given that tangent is parallel to the x-axis, so the slope of the tangent is equal to 0.

So,

$\frac{dy}{dx}=0$

or $0=3(x+1)(x-3)$

Thus, Either x = -1 or x = 3

When x = -1 we get y = 12 and if x =3 we get y = -20

So the required points are (-1, 12) and (3, -20).

Question:8 Find a point on the curve $y=(x-2{)}^2$ at which the tangent is parallel to the chord joining the points (2, 0) and

(4, 4).

Answer:

Points joining the chord is (2,0) and (4,4)
Now, we know that the slope of the curve with given two points is
$m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{4-2}=\frac{4}{2}=2$
As it is given that the tangent is parallel to the chord, so their slopes are equal
i.e. slope of the tangent = slope of the chord
Given the equation of the curve is $y=(x-2{)}^2$
$\therefore \frac{dy}{dx}=2(x-2)=2$
$$\begin{gathered} (x-2)=1 \\ x=1+2 \\ x=3 \end{gathered}$$
Now, when $x=3$ $y=(3-2{)}^2=(1{)}^2=1$
Hence, the coordinates are (3, 1)

Question:9 Find the point on the curve $y=x^3-11x+5$ at which the tangent is $y=x-11$

Answer:

We know that the equation of a line is y = mx + c
Know the given equation of tangent is
y = x - 11
So, by comparing with the standard equation we can say that the slope of the tangent (m) = 1 and value of c if -11
As we know that slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of curve is
$y=x^3-11x+5$
$\frac{dy}{dx}=3x^2-11$
$$\begin{gathered} 3x^2-11=1 \\ 3x^2=12 \\ {x}^2=4 \\ x=\pm 2 \end{gathered}$$
When x = 2 , $y=2^3-11(2)+5=8-22+5=-9$
and
When x = -2 , $y=(-2{)}^3-11(22)+5=-8+22+5=19$
Hence, the coordinates are (2,-9) and (-2,19), here (-2,19) does not satisfy the equation y=x-11

Hence, the coordinate is (2,-9) at which the tangent is $y=x-11$

Question:10 Find the equation of all lines having slope –1 that are tangents to the curve $y=\frac{1}{x-1},x\ne 1$

Answer:

We know that the slope of the tangent of at the point of the given curve is given by $\frac{dy}{dx}$

Given the equation of curve is
$y=\frac{1}{x-1}$
$\frac{dy}{dx}=\frac{-1}{(1-x{)}^2}$
It is given thta slope is -1
So,
$$\begin{gathered} \frac{-1}{(1-x{)}^2}=-1\Rightarrow (1-x{)}^2=1=1-x=\pm 1 \\ x=0andx=2 \end{gathered}$$
Now, when x = 0 , $y=\frac{1}{x-1}=\frac{1}{0-1}=-1$
and
when x = 2 , $y=\frac{1}{x-1}=\frac{1}{(2-1)}=1$
Hence, the coordinates are (0,-1) and (2,1)
Equation of line passing through (0,-1) and having slope = -1 is
y = mx + c
-1 = 0 X -1 + c
c = -1
Now equation of line is
y = -x -1
y + x + 1 = 0
Similarly, Equation of line passing through (2,1) and having slope = -1 is
y = mx + c
1 = -1 X 2 + c
c = 3
Now equation of line is
y = -x + 3
y + x - 3 = 0

Question:11 Find the equation of all lines having slope 2 which are tangents to the curve $y=\frac{1}{x-3},x\ne 3$

Answer:

We know that the slope of the tangent of at the point of the given curve is given by $\frac{dy}{dx}$

Given the equation of curve is
$y=\frac{1}{x-3}$
$\frac{dy}{dx}=\frac{-1}{(x-3{)}^2}$
It is given that slope is 2
So,
$\frac{-1}{(x-3{)}^2}=2\Rightarrow (x-3{)}^2=\frac{-1}{2}=x-3=\pm \frac{\sqrt{-}1}{\sqrt{2}}$
So, this is not possible as our coordinates are imaginary numbers
Hence, no tangent is possible with slope 2 to the curve $y=\frac{1}{x-3}$

Question:12 Find the equations of all lines having slope 0 which are tangent to the curve
$y=\frac{1}{x^2-2x+3}$

Answer:

We know that the slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$

Given the equation of the curve as
$y=\frac{1}{x^2-2x+3}$
$\frac{dy}{dx}=\frac{-(2x-2)}{(x^2-2x+3{)}^2}$
It is given thta slope is 0
So,
$\frac{-(2x-2)}{(x^2-2x+3{)}^2}=0\Rightarrow 2x-2=0=x=1$
Now, when x = 1 , $y=\frac{1}{x^2-2x+3}=\frac{1}{1^2-2(1)+3}=\frac{1}{1-2+3}=\frac{1}{2}$

Hence, the coordinates are $(1,\frac{1}{2})$
Equation of line passing through $(1,\frac{1}{2})$ and having slope = 0 is
y = mx + c
1/2 = 0 X 1 + c
c = 1/2
Now equation of line is
$y=\frac{1}{2}$

Question:13(i) Find points on the curve $\frac{x^2}{9}+\frac{y^2}{16}=1$ at which the tangents are parallel to x-axis

Answer:

Parallel to x-axis means slope of tangent is 0
We know that slope of tangent at a given point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve is
$\frac{x^2}{9}+\frac{y^2}{16}=1\Rightarrow 9y^2=144(1-16x^2)$
$18y\frac{dy}{dx}=-32x$
$\frac{dy}{dx}=\frac{(-32x)}{18y}=0\Rightarrow x=0$
From this, we can say that $x=0$
Now. when $x=0$ , $\frac{0^2}{9}+\frac{y^2}{16}=1\Rightarrow \frac{y^2}{16}=1\Rightarrow y=\pm 4$
Hence, the coordinates are (0,4) and (0,-4)

Question:13(ii) Find points on the curve $\frac{x^2}{9}+\frac{y^2}{16}=1$ at which the tangents are parallel to y-axis

Answer:

Parallel to y-axis means the slope of the tangent is $\mathrm{\infty }$ , means the slope of normal is 0
We know that slope of the tangent at a given point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve is
$\frac{x^2}{9}+\frac{y^2}{16}=1\Rightarrow 9y^2=144(1-16x^2)$
$18y\frac{dy}{dx}=144(1-32x)$
$\frac{dy}{dx}=\frac{-32x}{18y}=\mathrm{\infty }$
Slope of normal = $-\frac{dx}{dy}=\frac{18y}{32x}=0$
From this we can say that y = 0
Now. when y = 0, $\frac{x^2}{9}+\frac{0^2}{16}\Rightarrow 1=x=\pm 3$
Hence, the coordinates are (3,0) and (-3,0)

Question:14(i) Find the equations of the tangent and normal to the given curves at the indicated
points:
$y=x^4-6x^3+13x^2-10x+5at(0,5)$

Answer:

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$y=x^4-6x^3+13x^2-10x+5$
$\frac{dy}{dx}=4x^3-18x^2+26x-10$
at point (0,5)
$\frac{dy}{dx}=4(0{)}^3-18(0{)}^2+26(0)-10=-10$
Hence slope of tangent is -10
Now we know that,
$slopeofnormal=\frac{-1}{slopeoftangent}=\frac{-1}{-10}=\frac{1}{10}$
Now, equation of tangent at point (0,5) with slope = -10 is
$$\begin{gathered} y=mx+c \\ 5=0+c \\ c=5 \end{gathered}$$
equation of tangent is
$$\begin{gathered} y=-10x+5 \\ y+10x=5 \end{gathered}$$
Similarly, the equation of normal at point (0,5) with slope = 1/10 is
$$\begin{gathered} y=mx+c \\ 5=0+c \\ c=5 \end{gathered}$$
equation of normal is
$$\begin{gathered} y=\frac{1}{10}x+5 \\ 10y-x=50 \end{gathered}$$

Question:14(ii) Find the equations of the tangent and normal to the given curves at the indicated
points:
$y=x^4-6x^3+13x^2-10x+5at(1,3)$

Answer:

We know that Slope of tangent at a point on given curve is given by $\frac{dy}{dx}$
Given equation of curve
$y=x^4-6x^3+13x^2-10x+5$
$\frac{dy}{dx}=4x^3-18x^2+26x-10$
at point (1,3)
$\frac{dy}{dx}=4(1{)}^3-18(1{)}^2+26(1)-10=2$
Hence slope of tangent is 2
Now we know that,
$slopeofnormal=\frac{-1}{slopeoftangent}=\frac{-1}{2}$
Now, equation of tangent at point (1,3) with slope = 2 is
y = 2x + 1
y -2x = 1
Similarly, equation of normal at point (1,3) with slope = -1/2 is
y = mx + c
$3=\frac{-1}{2}\times 1+c$
$c=\frac{7}{2}$
equation of normal is
$$\begin{gathered} y=\frac{-1}{2}x+\frac{7}{2} \\ 2y+x=7 \end{gathered}$$

Question:14(iii) Find the equations of the tangent and normal to the given curves at the indicated
points:

$y=x^{3}at(1,1)$

Answer:

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$y=x^3$
$\frac{dy}{dx}=3x^2$
at point (1,1)
$\frac{dy}{dx}=3(1{)}^2=3$
Hence slope of tangent is 3
Now we know that,
$slopeofnormal=\frac{-1}{slopeoftangent}=\frac{-1}{3}$
Now, equation of tangent at point (1,1) with slope = 3 is
$$\begin{gathered} y=mx+c \\ 1=1\times 3+c \\ c=1-3=-2 \end{gathered}$$
equation of tangent is
$y-3x+2=0$
Similarly, equation of normal at point (1,1) with slope = -1/3 is
y = mx + c
$1=\frac{-1}{3}\times 1+c$
$c=\frac{4}{3}$
equation of normal is
$$\begin{gathered} y=\frac{-1}{3}x+\frac{4}{3} \\ 3y+x=4 \end{gathered}$$

Question:14(iv) Find the equations of the tangent and normal to the given curves at the indicated points

$y=x^{2}at(0,0)$

Answer:

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$y=x^2$
$\frac{dy}{dx}=2x$
at point (0,0)
$\frac{dy}{dx}=2(0{)}^2=0$
Hence slope of tangent is 0
Now we know that,
$slopeofnormal=\frac{-1}{slopeoftangent}=\frac{-1}{0}=-\mathrm{\infty }$
Now, equation of tangent at point (0,0) with slope = 0 is
y = 0
Similarly, equation of normal at point (0,0) with slope = $-\mathrm{\infty }$ is

$$\begin{gathered} y=x\times -\mathrm{\infty }+0 \\ x=\frac{y}{-\mathrm{\infty }} \\ x=0 \end{gathered}$$

Question:14(v) Find the equations of the tangent and normal to the given curves at the indicated points:

$x=\cos t,y=\sin tatt=\pi /4$

Answer:

We know that Slope of the tangent at a point on the given curve is given by $\frac{dy}{dx}$
Given the equation of the curve
$x=\cos t,y=\sin t$
Now,
$\frac{dx}{dt}=-\sin t$ and $\frac{dy}{dt}=\cos t$
Now,
${\left(\frac{dy}{dx}\right)}_{t=\frac{\pi }{4}}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\cos t}{-\sin t}=-\cot t==-\cot \frac{\pi }{4}=-1$
Hence slope of the tangent is -1
Now we know that,
$slopeofnormal=\frac{-1}{slopeoftangent}=\frac{-1}{-1}=1$
Now, the equation of the tangent at the point $t=\frac{\pi }{4}$ with slope = -1 is
$x=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ and

$y=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$
equation of the tangent at

$t=\frac{\pi }{4}$ i.e. $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ is

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-\frac{1}{\sqrt{2}}=-1(x-\frac{1}{\sqrt{2}}) \\ \sqrt{2}y+\sqrt{2}x=2 \\ y+x=\sqrt{2} \end{gathered}$$
Similarly, the equation of normal at $t=\frac{\pi }{4}$ with slope = 1 is
$x=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$ and

$y=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$
equation of the tangent at

$t=\frac{\pi }{4}$ i.e. $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ is
$$\begin{gathered} y-y_1=m(x-x_1) \\ y-\frac{1}{\sqrt{2}}=1(x-\frac{1}{\sqrt{2}}) \\ \sqrt{2}y-\sqrt{2}x=0 \\ y-x=0 \\ x=y \end{gathered}$$

Question:15(a) Find the equation of the tangent line to the curve $y=x^2-2x+7$ which is parallel to the line $2x-y+9=0$

Answer:

Parellel to line $2x-y+9=0$ means slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get slope of line m = 2 and c = 9
Now, we know that the slope of tangent at a given point to given curve is given by $\frac{dy}{dx}$
Given equation of curve is
$y=x^2-2x+7$
$$\begin{gathered} \frac{dy}{dx}=2x-2=2 \\ x=2 \end{gathered}$$
Now, when x = 2 , $y=(2{)}^2-2(2)+7=4-4+7=7$
Hence, the coordinates are (2,7)
Now, equation of tangent paasing through (2,7) and with slope m = 2 is
y = mx + c
7 = 2 X 2 + c
c = 7 - 4 = 3
So,
y = 2 X x+ 3
y = 2x + 3
So, the equation of tangent is y - 2x = 3

Question:15(b) Find the equation of the tangent line to the curve $y=x^2-2x+7$ which is perpendicular to the line $5y-15x=13.$

Answer:

Perpendicular to line $5y-15x=13.\Rightarrow y=3x+\frac{13}{5}$ means $slopeoftangent=\frac{-1}{slopeofline}$
We know that the equation of the line is
y = mx + c
on comparing with the given equation we get the slope of line m = 3 and c = 13/5
$slopeoftangent=\frac{-1}{slopeofline}=\frac{-1}{3}$
Now, we know that the slope of the tangent at a given point to given curve is given by $\frac{dy}{dx}$
Given the equation of curve is
$y=x^2-2x+7$
$$\begin{gathered} \frac{dy}{dx}=2x-2=\frac{-1}{3} \\ x=\frac{5}{6} \end{gathered}$$
Now, when $x=\frac{5}{6}$ , $y=(\frac{5}{6}{)}^2-2(\frac{5}{6})+7=\frac{25}{36}-\frac{10}{6}+7=\frac{217}{36}$
Hence, the coordinates are $(\frac{5}{6},\frac{217}{36})$
Now, the equation of tangent passing through (2,7) and with slope $m=\frac{-1}{3}$ is
$$\begin{gathered} y=mx+c \\ \frac{217}{36}=\frac{-1}{3}\times \frac{5}{6}+c \\ c=\frac{227}{36} \end{gathered}$$
So,
$$\begin{gathered} y=\frac{-1}{3}x+\frac{227}{36} \\ 36y+12x=227 \end{gathered}$$
Hence, equation of tangent is 36y + 12x = 227

Question:16 Show that the tangents to the curve $y=7x^3+11$ at the points where x = 2 and x = – 2 are parallel .

Answer:

Slope of tangent = $\frac{dy}{dx}=21x^2$
When x = 2
$\frac{dy}{dx}=21x^2=21(2{)}^2=21\times 4=84$
When x = -2
$\frac{dy}{dx}=21x^2=21(-2{)}^2=21\times 4=84$
Slope is equal when x= 2 and x = - 2
Hence, we can say that both the tangents to curve $y=7x^3+11$ is parallel

Question:17 Find the points on the curve $y=x^3$ at which the slope of the tangent is equal to the y-coordinate of the point.

Answer:

Given equation of curve is $y=x^3$
Slope of tangent = $\frac{dy}{dx}=3x^2$
it is given that the slope of the tangent is equal to the y-coordinate of the point
$3x^2=y$
We have $y=x^3$
$$\begin{gathered} 3x^2=x^3 \\ 3x^2-x^3=0 \\ {x}^2(3-x)=0 \\ x=0andx=3 \end{gathered}$$
So, when x = 0 , y = 0
and when x = 3 , $y=x^3=3^3=27$

Hence, the coordinates are (3,27) and (0,0)

Question:18 For the curve $y=4x^3-2x^5$ , find all the points at which the tangent passes
through the origin.

Answer:

Tangent passes through origin so, (x,y) = (0,0)
Given equtaion of curve is $y=4x^3-2x^5$
Slope of tangent =

$\frac{dy}{dx}=12x^2-10x^4$
Now, equation of tangent is
$Y-y=m(X-x)$
at (0,0) Y = 0 and X = 0
$-y=(12x^3-10x^4)(-x)$
$y=12x^3-10x^5$
and we have $y=4x^3-2x^5$
$4x^3-2x^5=12x^3-10x^5$
$$\begin{gathered} 8x^5-8x^3=0 \\ 8x^3(x^2-1)=0 \\ x=0andx=\pm 1 \end{gathered}$$
Now, when x = 0,

$y=4(0{)}^3-2(0{)}^5=0$
when x = 1 ,

$y=4(1{)}^3-2(1{)}^5=4-2=2$
when x= -1 ,

$y=4(-1{)}^3-2(-1{)}^5=-4-(-2)=-4+2=-2$
Hence, the coordinates are (0,0) , (1,2) and (-1,-2)

Question:19 Find the points on the curve $x^2+y^2-2x-3=0$ at which the tangents are parallel
to the x-axis.

Answer:

parellel to x-axis means slope is 0
Given equation of curve is
$x^2+y^2-2x-3=0$
Slope of tangent =
$$\begin{gathered} -2y\frac{dy}{dx}=2x-2 \\ \frac{dy}{dx}=\frac{1-x}{y}=0 \\ x=1 \end{gathered}$$
When x = 1 ,

$-y^2=x^2-2x-3=(1{)}^2-2(1)-3=1-5=-4$
$y=\pm 2$
Hence, the coordinates are (1,2) and (1,-2)

Question:20 Find the equation of the normal at the point $(a{m}^2,a{m}^3)$ for the curve $a{y}^2=x^3.$

Answer:

Given equation of curve is
$a{y}^2=x^3\Rightarrow y^2=\frac{x^3}{a}$
Slope of tangent

$2y\frac{dy}{dx}=\frac{3x^2}{a}\Rightarrow \frac{dy}{dx}=\frac{3x^2}{2ya}$
at point $(a{m}^2,a{m}^3)$
$\frac{dy}{dx}=\frac{3(a{m}^2{)}^2}{2(a{m}^3)a}=\frac{3a^2{m}^4}{2a^2{m}^3}=\frac{3m}{2}$
Now, we know that
$Slopeofnormal=\frac{-1}{Slopeoftangent}=\frac{-2}{3m}$
equation of normal at point $(a{m}^2,a{m}^3)$ and with slope $\frac{-2}{3m}$
$$\begin{gathered} y-y_1=m(x-x_1) \\ y-a{m}^3=\frac{-2}{3m}(x-a{m}^2) \\ 3ym-3a{m}^4=-2(x-a{m}^2) \\ 3ym+2x=3a{m}^4+2a{m}^2 \end{gathered}$$
Hence, the equation of normal is $3ym+2x=3a{m}^4+2a{m}^2$

Question:21 Find the equation of the normals to the curve $y=x^3+2x+6$ which are parallel
to the line $x+14y+4=0.$

Answer:

Equation of given curve is
$y=x^3+2x+6$
Parellel to line $x+14y+4=0\Rightarrow y=\frac{-x}{14}-\frac{4}{14}$ means slope of normal and line is equal
We know that, equation of line
y= mx + c
on comparing it with our given equation. we get,
$m=\frac{-1}{14}$
Slope of tangent = $\frac{dy}{dx}=3x^2+2$
We know that
$Slopeofnormal=\frac{-1}{Slopeoftangent}=\frac{-1}{3x^2+2}$
$\frac{-1}{3x^2+2}=\frac{-1}{14}$
$$\begin{gathered} 3x^2+2=14 \\ 3x^2=12 \\ {x}^2=4 \\ x=\pm 2 \end{gathered}$$
Now, when x = 2, $y=(2{)}^3+2(2)+6=8+4+6=18$
and
When x = -2 , $y=(-2{)}^3+2(-2)+6=-8-4+6=-6$
Hence, the coordinates are (2,18) and (-2,-6)
Now, the equation of at point (2,18) with slope $\frac{-1}{14}$
$$\begin{gathered} y-y_1=m(x-x_1) \\ y-18=\frac{-1}{14}(x-2) \\ 14y-252=-x+2 \\ x+14y=254 \end{gathered}$$
Similarly, the equation of at point (-2,-6) with slope $\frac{-1}{14}$

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-(-6)=\frac{-1}{14}(x-(-2)) \\ 14y+84=-x-2 \\ x+14y+86=0 \end{gathered}$$
Hence, the equation of the normals to the curve $y=x^3+2x+6$ which are parallel
to the line $x+14y+4=0.$

are x +14y - 254 = 0 and x + 14y +86 = 0

Question:22 Find the equations of the tangent and normal to the parabola $y^2=4ax$ at the point $(a{t}^2,2at).$

Answer:

Equation of the given curve is
$y^2=4ax$

Slope of tangent = $2y\frac{dy}{dx}=4a\Rightarrow \frac{dy}{dx}=\frac{4a}{2y}$
at point $(a{t}^2,2at).$
$\frac{dy}{dx}=\frac{4a}{2(2at)}=\frac{4a}{4at}=\frac{1}{t}$
Now, the equation of tangent with point $(a{t}^2,2at).$ and slope $\frac{1}{t}$ is
$$\begin{gathered} y-y_1=m(x-x_1) \\ y-2at=\frac{1}{t}(x-a{t}^2) \\ yt-2a{t}^2=x-a{t}^2 \\ x-yt+a{t}^2=0 \end{gathered}$$

We know that
$Slopeofnormal=\frac{-1}{Slopeoftangent}=-t$
Now, the equation of at point $(a{t}^2,2at).$ with slope -t
$$\begin{gathered} y-y_1=m(x-x_1) \\ y-2at=(-t)(x-a{t}^2) \\ y-2at=-xt+a{t}^3 \\ xt+y-2at-a{t}^3=0 \end{gathered}$$
Hence, the equations of the tangent and normal to the parabola

$y^2=4ax$ at the point $(a{t}^2,2at).$ are
$x-yt+a{t}^2=0andxt+y-2at-a{t}^3=0respectively$

Question:23 Prove that the curves $x=y^2$ and xy = k cut at right angles* $if8k^2=1.$

Answer:

Let suppose, Curve $x=y^2$ and xy = k cut at the right angle
then the slope of their tangent also cut at the right angle
means,
${\left(\frac{dy}{dx}\right)}_a\times {\left(\frac{dy}{dx}\right)}_b=-1$ -(i)
$2y{\left(\frac{dy}{dx}\right)}_a=1\Rightarrow {\left(\frac{dy}{dx}\right)}_a=\frac{1}{2y}$
${\left(\frac{dy}{dx}\right)}_b=\frac{-k}{x^2}$
Now these values in equation (i)
$$\begin{gathered} \frac{1}{2y}\times \frac{-k}{x^2}=-1 \\ -k=-2y{x}^2 \\ k=2(xy)(x) \\ k=2k(k^{\frac{2}{3}})(x=y^2\Rightarrow y^{2}y=k\Rightarrow y=k^{\frac{1}{3}}andx=k^{\frac{2}{3}}) \\ 2(k^{\frac{2}{3}})=1 \\ {(2(k^{\frac{2}{3}}))}^3=1^3 \\ 8k^2=1 \end{gathered}$$
Hence proved