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We have seen the graph of $\sin \theta$ or $\cos \theta$, and by that graph, we have observed that it attains the maximum and minimum value in a given interval of [0,$\frac{\pi}{2}$]. $\sin \theta$ obtains the maximum value at $\frac{\pi}{2}$ and the minimum value at 0. Similarly, $\cos \theta$ attains the maximum value at 0, and the minimum value at $\frac{\pi}{2}$. In this exercise, we will learn to find the local minima and maxima. Also, we will learn to find the absolute minima and maxima for a given function. Local maxima and minima refer to the highest and lowest points of a function within a specific interval or region, while absolute maxima and minima represent the highest and lowest points of the function across its entire domain. The NCERT solutions for Class 12 Maths chapter 6 exercise 6.3 use the concept of derivatives to find the maximum and minimum of different functions. In the NCERT Class 12 Mathematics Book, some real-life examples of finding maximum and minimum values are given.
Question:1 . Find the slope of the tangent to the curve $y = 3 x ^4 - 4x \: \: at \: \: x \: \: = 4$
Answer:
Given curve is,
$y = 3 x ^4 - 4x$
Now, the slope of the tangent at point x =4 is given by
$\left ( \frac{dy}{dx} \right )_{x=4} = 12x^3 - 4$
$= 12(4)^3-4$
$= 12(64)-4 = 768 - 4 =764$
Question:2 . Find the slope of the tangent to the curve $\frac{x-1}{x-2} , x \neq 2 \: \: at\: \: x = 10$
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
Answer:
The slope of the tangent at a point on a given curve is given by
$\left ( \frac{dy}{dx} \right )$
Now,
$\left ( \frac{dx}{d \theta} \right )_{\theta=\frac{\pi}{4}} = 3a\cos^2 \theta(-\sin \theta) = 3a(\frac{1}{\sqrt2})^2(-\frac{1}{\sqrt2}) = -\frac{3\sqrt2 a}{4}$
Similarly,
$\left ( \frac{dy}{d \theta} \right )_{\theta=\frac{\pi}{4}} = 3a\sin^2 \theta(\cos \theta) = 3a(\frac{1}{\sqrt2})^2(\frac{1}{\sqrt2}) = \frac{3\sqrt2 a}{4}$
$\left ( \frac{dy}{dx} \right ) = \frac{\left ( \frac{dy}{d\theta} \right )}{\left ( \frac{dx}{d\theta} \right )} = \frac{\frac{3\sqrt2 a}{4}}{-\frac{3\sqrt2 a}{4}} = -1$
Hence, the slope of the tangent at $\theta = \frac{\pi}{4}$ is -1
Now,
Slope of normal = $-\frac{1}{slope \ of \ tangent}$ = $-\frac{1}{-1} = 1$
Hence, the slope of normal at $\theta = \frac{\pi}{4}$ is 1
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,
$\left ( \frac{dy}{d \theta} \right )_{\theta=\frac{\pi}{2}} = 2b\cos \theta(-\sin \theta)$
$\left ( \frac{dy}{dx} \right )_{x=\frac{\pi}{2}} = \frac{\left ( \frac{dy}{d\theta} \right )}{\left ( \frac{dx}{d\theta} \right )} = \frac{-2b\cos \theta \sin \theta}{-a\cos \theta} = \frac{2b\sin \theta}{a} = \frac{2b\times1}{a} = \frac{2b}{a}$
Hence, the slope of the tangent at $\theta = \frac{\pi}{2}$ is $\frac{2b}{a}$
Now,
Slope of normal = $-\frac{1}{slope \ of \ tangent}$ = $-\frac{1}{\frac{2b}{a}} = -\frac{a}{2b}$
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 $y = x^3 - 3 x^2 - 9x +7$ is parallel to the x-axis.
Answer:
We are given :
$y = x^3 - 3 x^2 - 9x +7$
Differentiating the equation with respect to x, we get :
$\frac{dy}{dx}\ =\ 3x^2\ -\ 6x\ -\ 9\ +\ 0$
or $=\ 3\left ( x^2\ -\ 2x\ -\ 3 \right )$
or $\frac{dy}{dx}\ =\ 3\left ( x+1 \right )\left ( x-3 \right )$
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 $0\ =\ 3\left ( x+1 \right )\left ( x-3 \right )$
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).
(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 $y = ( x-2)^2$
$\therefore \frac{dy}{dx} = 2(x-2) = 2$
$(x-2) = 1\\ x = 1+2\\ x=3$
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
$y = x^3 - 11x + 5$
$\frac{dy}{dx} = 3x^2 -11$
$3x^2 -11 = 1\\ 3x^2 = 12 \\ x^2 = 4 \\ x = \pm2$
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 $y = \frac{1}{x-1} , x \neq 1$
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,
$\frac{-1}{(1-x)^2} = -1 \Rightarrow (1-x)^2 = 1 = 1 - x = \pm 1 \\ \\ x = 0 \ and \ x = 2$
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 $y = \frac{1}{x-3} , x \neq 3$
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,
$\frac{-1}{(x-3)^2} = 2 \Rightarrow (x-3)^2 = \frac{-1}{2} = x-3 = \pm \frac{\sqrt-1}{\sqrt2} \\ \\$
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
$y = \frac{1}{x^2 - 2 x +3 }$
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 $\left ( 1,\frac{1}{2} \right )$
Equation of line passing through $\left ( 1,\frac{1}{2} \right )$ 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 $\infty$ , 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
$\frac{x^2 }{9} + \frac{y^2 }{16} = 1 \Rightarrow 9y^2 = 144(1-16x^2)$
$18y\frac{dy}{dx} = 144(1-32x)$
$\frac{dy}{dx} = \frac{-32x}{18y} = \infty$
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)
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,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{-10} = \frac{1}{10}$
Now, equation of tangent at point (0,5) with slope = -10 is
$y = mx + c\\ 5 = 0 + c\\ c = 5$
equation of tangent is
$y = -10x + 5\\ y + 10x = 5$
Similarly, the equation of normal at point (0,5) with slope = 1/10 is
$\\y = mx + c \\5 = 0 + c \\c = 5$
equation of normal is
$\\y = \frac{1}{10}x+5 \\ 10y - x = 50$
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,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{2}$
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
$y = \frac{-1}{2}x+\frac{7}{2} \\ 2y + x = 7$
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,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{3}$
Now, equation of tangent at point (1,1) with slope = 3 is
$y = mx + c\\ 1 = 1 \times 3 + c\\ c = 1 - 3 = -2$
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
$y = \frac{-1}{3}x+\frac{4}{3} \\ 3y + x = 4$
Question:14(iv) Find the equations of the tangent and normal to the given curves at the indicated points
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,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{0} = -\infty$
Now, equation of tangent at point (0,0) with slope = 0 is
y = 0
Similarly, equation of normal at point (0,0) with slope = $-\infty$ is
$\\y = x \times -\infty + 0\\ x = \frac{y}{-\infty}\\ x=0$
Question:14(v) Find the equations of the tangent and normal to the given curves at the indicated points:
$x = \cos t , y = \sin t \: \: at \: \: t = \pi /4$
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,
$slope \ of \ normal = \frac{-1}{slope \ of \ tangent} = \frac{-1}{-1} = 1$
Now, the equation of the tangent at the point $t = \frac{\pi}{4}$ with slope = -1 is
$x= \cos \frac{\pi}{4} = \frac{1}{\sqrt2}$ and
$y= \sin \frac{\pi}{4} = \frac{1}{\sqrt2}$
equation of the tangent at
$t = \frac{\pi}{4}$ i.e. $\left ( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right )$ is
$y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = -1(x- \frac{1}{\sqrt2})\\ \sqrt2y + \sqrt2x = 2\\ y + x = \sqrt2$
Similarly, the equation of normal at $t = \frac{\pi}{4}$ with slope = 1 is
$x= \cos \frac{\pi}{4} = \frac{1}{\sqrt2}$ and
$y= \sin \frac{\pi}{4} = \frac{1}{\sqrt2}$
equation of the tangent at
$t = \frac{\pi}{4}$ i.e. $\left ( \frac{1}{\sqrt2}, \frac{1}{\sqrt2}\right )$ is
$\\y- y_1 = m(x-x_1)\\ y-\frac{1}{\sqrt2} = 1(x- \frac{1}{\sqrt2})\\ \sqrt2y - \sqrt2x = 0\\ y - x = 0\\ x=y$
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
$y = x^2 - 2x +7$
$\frac{dy}{dx} = 2x - 2 = 2\\ \\ x = 2$
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 $slope \ of \ tangent = \frac{-1}{slope \ of \ line}$
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
$slope \ of \ tangent = \frac{-1}{slope \ of \ line} = \frac{-1}{3}$
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
$y = x^2 - 2x +7$
$\frac{dy}{dx} = 2x - 2 = \frac{-1}{3}\\ \\ x = \frac{5}{6}$
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
$y = mx+ c\\ \frac{217}{36}= \frac{-1}{3}\times \frac{5}{6} + c\\ c = \frac{227}{36}$
So,
$y = \frac{-1}{3}x+\frac{227}{36}\\ 36y + 12x = 227$
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
$\frac{dy}{dx} = 21x^2 = 21(2)^{2} = 21 \times4 = 84$
When x = -2
$\frac{dy}{dx} = 21x^2 = 21(-2)^{2} = 21 \times4 = 84$
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
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 $y = x ^3$
$3x^2 = x^3\\ 3x^2 - x^3=0\\ x^2(3-x)=0\\ x= 0 \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ x = 3$
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 $y = 4x ^ 3 - 2x ^5$
$4x^3-2x^5= 12x^3-10x^5$
$8x^5 - 8x^3=0\\ 8x^3(x^2-1)=0\\ x=0\ \ \ \ \ \ and \ \ \ \ \ \ \ x = \pm1$
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 =
$-2y\frac{dy}{dx} = 2x -2\\ \frac{dy}{dx} = \frac{1-x}{y} = 0\\ x= 1$
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 $( am^2 , am^3 )$ for the curve $ay ^2 = x ^3.$
Answer:
Given equation of curve is
$ay ^2 = x ^3\Rightarrow y^2 = \frac{x^3}{a}$
Slope of tangent
$2y\frac{dy}{dx} = \frac{3x^2 }{a} \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2ya}$
at point $( am^2 , am^3 )$
$\frac{dy}{dx} = \frac{3(am^2)^2}{2(am^3)a} = \frac{3a^2m^4}{2a^2m^3} = \frac{3m}{2}$
Now, we know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-2}{3m}$
equation of normal at point $( am^2 , am^3 )$ and with slope $\frac{-2}{3m}$
$y-y_1=m(x-x_1)\\ y-am^3 = \frac{-2}{3m}(x-am^2)\\ 3ym - 3am^4 = -2(x-am^2)\\ 3ym +2x= 3am^4+2am^2$
Hence, the equation of normal is $3ym +2x= 3am^4+2am^2$
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
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-1}{3x^2+2}$
$\frac{-1}{3x^2+2} = \frac{-1}{14}$
$3x^2+2 = 14\\ 3x^2 = 12 \\ x^2 = 4\\ x = \pm 2$
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 $\frac{-1}{14}$
$y-y_1=m(x-x_1)\\ y-18=\frac{-1}{14}(x-2)\\ 14y - 252 = -x + 2\\ x+14y = 254$
Similarly, the equation of at point (-2,-6) with slope $\frac{-1}{14}$
$y-y_1=m(x-x_1)\\ y-(-6)=\frac{-1}{14}(x-(-2))\\ 14y + 84 = -x - 2\\ x+14y + 86= 0$
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 $y ^2 = 4 ax$ at the point $(at ^2, 2at).$
Answer:
Equation of the given curve is
$y ^2 = 4 ax$
Slope of tangent = $2y\frac{dy}{dx} = 4a \Rightarrow \frac{dy}{dx} = \frac{4a}{2y}$
at point $(at ^2, 2at).$
$\frac{dy}{dx}= \frac{4a}{2(2at)} = \frac{4a}{4at} = \frac{1}{t}$
Now, the equation of tangent with point $(at ^2, 2at).$ and slope $\frac{1}{t}$ is
$y-y_1=m(x-x_1)\\ y-2at=\frac{1}{t}(x-at^2)\\ yt - 2at^2 = x - at^2\\ x-yt +at^2 = 0$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = -t$
Now, the equation of at point $(at ^2, 2at).$ with slope -t
$y-y_1=m(x-x_1)\\ y-2at=(-t)(x-at^2)\\ y - 2at = -xt + at^3\\ xt+y -2at -at^3 = 0$
Hence, the equations of the tangent and normal to the parabola
$y ^2 = 4 ax$ at the point $(at ^2, 2at).$ are
$x-yt+at^2=0\ \ \ \ and \ \ \ \ xt+y -2at -at^3 = 0 \ \ respectively$
Question:23 Prove that the curves $x = y^2$ and xy = k cut at right angles* $if \: \: 8k ^ 2 = 1.$
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)
$\frac{1}{2y} \times \frac{-k}{x^2} = -1\\ -k = -2yx^2\\ k =2(xy)(x)\\ k = 2k(k^{\frac{2}{3}}) \ \ \ \ \left ( x = y^2 \Rightarrow y^2y = k \Rightarrow y = k^{\frac{1}{3}} \ and \ x = k^{\frac{2}{3}} \right ) \\ 2(k^{\frac{2}{3}}) = 1\\ \left ( 2(k^{\frac{2}{3}}) \right )^3 = 1^3\\ 8k^2 = 1$
Hence proved
Answer:
Given equation is
$\frac{x^2 }{a^2} - \frac{y^2 }{b^2 }= 1 \Rightarrow y^2a^2 = x^2b^2 -a^2b^2$
Now ,we know that
slope of tangent = $2ya^2\frac{dy}{dx} = 2xb^2 \Rightarrow \frac{dy}{dx} = \frac{xb^2}{ya^2}$
at point $(x_0 , y_0 )$
$\frac{dy}{dx} = \frac{x_0b^2}{y_0a^2}$
equation of tangent at point $(x_0 , y_0 )$ with slope $\frac{xb^2}{ya^2}$
$y-y_1=m(x-x_1)\\ y-y_0=\frac{x_0b^2}{y_0a^2}(x-x_0)\\ yy_0a^2-y_0^2a^2 = xx_0b^2-x_0^2b^2\\ xx_0b^2 - yy_0a^2 = x_0^2b^2-y_0^2a^2$
Now, divide both sides by $a^2b^2$
$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = \left ( \frac{x_0^2}{a^2} - \frac{y_0^2}{b^2} \right )$
$=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left ( \because \frac{x_0^2}{a^2}-\frac{y_0^2}{b^2 } = 1\right )$
$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$
Hence, the equation of tangent is
$\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$
We know that
$Slope \ of \ normal= \frac{-1}{slope \ of \ tangent } = -\frac{y_0a^2}{x_0b^2}$
equation of normal at the point $(x_0 , y_0 )$ with slope $-\frac{y_0a^2}{x_0b^2}$
$y-y_1=m(x-x_1)\\ y-y_0=-\frac{y_0a^2}{x_0b^2}(x-x_0)\\ \frac{y-y_0}{y_0a^2} + \frac{x-x_0}{x_0b^2} = 0$
Answer:
Parellel to line $4x - 2y + 5 = 0 \Rightarrow y = 2x + \frac{5}{2}$ means the 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 the slope of line m = 2 and c = 5/2
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
$y = \sqrt{3x-2}$
$\frac{dy}{dx} = \frac{1}{2}.\frac{3}{\sqrt{3x-2}}=\frac{3}{2\sqrt{3x-2}}$
$\frac{3}{2\sqrt{3x-2}} = 2\\ 3^2 = (4\sqrt{3x-2})^2\\ 9 = 16(3x-2)\\ 3x-2=\frac{9}{16}\\ 3x = \frac{9}{16} +2\\ 3x= \frac{41}{16}\\ x = \frac{41}{48}$
Now, when
$x = \frac{41}{48}$ , $y = \sqrt{3x-2} \Rightarrow y = \sqrt{3\times\frac{41}{48}-2 } = \sqrt{\frac{41}{16}-2}=\sqrt\frac{9}{16 } = \pm \frac{3}{4}$
but y cannot be -ve so we take only positive value
Hence, the coordinates are
$\left ( \frac{41}{48},\frac{3}{4} \right )$
Now, equation of tangent paasing through
$\left ( \frac{41}{48},\frac{3}{4} \right )$ and with slope m = 2 is
$y - y_1=m(x-x_1)\\ y-\frac{3}{4}=2(x-\frac{41}{48})\\ 48y-36=2(48x-41)\\ 48x-24y=41-18\\ 48x-24y=23$
Hence, equation of tangent paasing through $\left ( \frac{41}{48},\frac{3}{4} \right )$ and with slope m = 2 is 48x - 24y = 23
Answer:
Equation of the given curve is
$y = 2x ^2 + 3 \sin x$
Slope of tangent = $\frac{dy}{dx} = 4x +3 \cos x$
at x = 0
$\frac{dy}{dx} = 4(0) +3 \cos 0= 0 + 3$
$\frac{dy}{dx}= 3$
Now, we know that
$Slope \ of \ normal = \frac{-1}{\ Slope \ of \ tangent} = \frac{-1}{3}$
Hence, (D) is the correct option
Answer:
The slope of the given line $y = x+1$ is 1
given curve equation is
$y^2 = 4 x$
If the line is tangent to the given curve than the slope of the tangent is equal to the slope of the curve
The slope of tangent = $2y\frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$
$\frac{dy}{dx} = \frac{2}{y} = 1\\ y = 2$
Now, when y = 2, $x = \frac{y^2}{4} = \frac{2^2}{4} = \frac{4}{4} = 1$
Hence, the coordinates are (1,2)
Hence, (A) is the correct answer
Also Read,
Maxima and Minima:
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Maximum and Minimum Values of a Function in a Closed Interval:
When we are given a continuous function $f(x)$ on a closed interval $[a, b]$, the function must attain a maximum and a minimum value somewhere in that interval.
These are called the absolute maximum and absolute minimum.
Also, read,
These are links to other subjects' NCERT textbook solutions. Students can check and analyse these well-structured solutions for a deeper understanding.
Students can check these NCERT exemplar links for further practice purposes.
Frequently Asked Questions (FAQs)
y-y0=f’(x0)(x-x0)
(y-y0)f’(x0)+(x-x0)=0
The slope =0. Therefore the equation of tangent is y=y0
The equation of tangent with infinite slope is x=x0
Normal is perpendicular to the tangent
The slope of normal is the negative of the inverse of the slope of the tangent.
27 questions are answered in the NCERT Solutions for Class 12 Maths chapter 6 exercise 6.3. For more questions refer to NCERT exemplar. Following NCERT syllabus will be useful for CBSE board exams.
There are 7 solved examples explained before the NCERT book Class 12 Maths chapter 6 exercise 6.3.
On Question asked by student community
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Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.
From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .
If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.
The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.
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