Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives -Exercise 6.3 | Set 2

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

(i) y = x⁴– 6x³+ 13x²– 10x+ 5 at (0, 5)

(ii) y = x⁴– 6x³+ 13x²– 10x + 5 at (1, 3)

(iii) y = x³ at(1, 1)

(iv) y = x² at(0, 0)

(v) x = cos t, y = sin t at t = π/4

Solution:

(i) Given curve
y = x⁴– 6x³+ 13x²– 10x + 5
Given point, (0, 5)
dy/dx = 4x³– 18x²+ 26x – 10
dy/dx = 4(0)³– 18(0)²+ 26(0) – 10
dy/dx = -10,
-dx/dy = 1/10
Now, with the help of points slope form
y – y₁= m(x – x₁)
y – 5 = -10(x – 0)
y + 10x = 5 is the required equation of the tangent
For equation of normal,
y – y1 = $\frac{-dx}{dy}(x-x_1)$
y – 5 = $\frac{1}{10}(x-0)$
10y – x – 50 is the equation of normal.
(ii) Given curve: y = x⁴– 6x³+ 13x²– 10x + 5
Given point is (1, 3)
dy/dx = 4x³– 18x²+ 26x – 10
dy/dx = 4(1)³– 18(1)²+ 26(1) – 10
dy/dx = 4 – 18 + 26 – 10 = 2
dy/dx = 2
-dx/dy = -1/2
Using point slope form, equation of tangent is
y – y₁= dy/dx(x – x1)
y – 3 = 2(x – 1)
y – 2x = 1 is the equation of tangent.
Using point slope form, equation of normal is
y – y₁= -dx/dy(x – 1)
y – 3 = -1/2(x – 1)
2y – 6 = -x + 1
2y + x = 7 is the equation of normal.
(iii) Given curve : y = x³
Given point is (1, 1)
dy/dx = 3x²
dy/dx = 3(1)²= 3
dy/dx = 3 & -dx/dy = -1/3
Using point slope form, equation of tangent is y – y₁= dy/dx(x – x₁)
y – y₁= dy/dx(x – x₁)
y – 1 = 3(x – 1)
y – 3x + 2 = 0 is the equation of tangent
Using point slope form, equation of normal is
y – y1 = $\frac{-dx}{dy}(x-x_1)$
y – 1 = $\frac{-1}{3}(x-1)$
3y – 3 = -x + 1
3y + x = 4 is the equation of normal.
(iv) Given curve: y = x²
Given point (0, 0)
dy/dx = 2x
dy/dx = 0
dy/dx = 0 & -dx/dy = not defined is
y – y₁= dy/dx(x – x₁)
y – 0 = 0(x – 0)
y = 0 is the equation of tangent.
Using point slope form, equation of normal is
y – y₁= $\frac{-dx}{dy}(x-x_1)$ -(1)
-dx\dy is undefined, so we can write eq(1) as
$\frac{-dy}{dx}(y-y_1)=x-x_1$
Now putting dy/dx = 0 we get
0(y – 0) = x-0
x = 0 is the equation of normal.
(v) Equation of curve: x = cos t and y = sin t
Point t = π/4
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ -(1)
$\frac{dx}{dt}=\frac{d(\cos t)}{dt}=-\sin t$
$\frac{dy}{dt}=\frac{d(\sin t)}{dt}=\cos t$
On putting these values in eq(1), we get
$\frac{dy}{dx}=\frac{\cos t}{-\sin t}=-\cot t$
$\frac{dy}{dx}=-\cot \frac{π}{4}=-1$
dy/dx = -1 & -dx/dy = 1
Now for t = π/4,
y₁= sin t = sin(π/4) = 1/√2
x₁= cos t = cos(π/4) = 1/√2
The point is (1/√2, 1/√2)
y – y₁= $\frac{dy}{dx}(x-x_1)$
y – (1/√2) = -1(x – 1/√2)
y – 1/√2 = -x + 1/√2
x + y = √2 is the equation of normal is
$y-y_1=\frac{-dx}{dy}(x-x_1)$
y – 1/√2 = 1(x – 1/√2)
x = y is the equation of normal.

Question 15. Find the equation of the tangent line to the curve y = x²– 2x + 7 which is

(i) Parallel to line 2x – y + 9 = 0

(ii) Perpendicular to the line 5y – 15x = 13

Solution:

Given curve: y = x²– 2x + 7
On differentiating w.r.t. x, we get
dy/dx = 2x – 2 -(1)
(i) Tangent is parallel to 2x – y + 9 = 0 that means,
Slope of tangent = slope of 2x – y + 9 = 0
y = 2x + 9
Slope = 2 -(Comparing with y = mx + e)
dy/dx = slope = 2
2x – 2 = 2
x₁= 2
Corresponding to x₁= 2,
y₁ = x₁² – 2x₁ + 7
y₁= (2)²– 2(2) + 7
y₁= 7
The point of contact is (2, 7).
Using point slope form, equation of tangent is
y – y₁= $\frac{dy}{dx}(x-x_1)$
y – 7 = 2(x – 2)
y – 2x = 3 is the equation of tangent.
(ii) Tangent is perpendicular to the line 5y – 15x = 13
That means (slope of tangent) x (slope of line) = -1
For, slope of line 5y – 15x = 13
5y = 15x + 13
y = 3x + 13/15
Slope = 3
(Slope of tangent) x 3 = -1
Slope of tangent =-1/3
Now, y = x²– 2x + 7
dy/dx = 2x – 2
On comparing dy/dx with slope, we get
2x – 2 = -1/3
6x – 6 = -1
6x = 5
x₁ = 5/6
For x₁= 5/6,
y₁= x₁² – 2x₁ + 7
y₁= (5/6)₁² – 2(5/6)₁ + 7
y₁= 217/36
Now using point slope form, equation of tangent is
y – y₁ = m(x – x₁)
$y-\frac{217}{36}=\frac{-1}{3}(x-\frac{5}{6})$
$\frac{36y-217}{36}=-1\frac{(6x-5)}{6.3}$
36y – 217 = -12x + 10
36y + 12x = 227 is the required equation of tangent.

Question 16. Show that the tangent to the curve y = 7x³+ 11 at the points where x = 2 and x = -2 are parallel.

Solution:

Given curve: y = 7x³+ 11
On differentiating w.r.t. x, we get
dy/dx = 21x²
dy/dx = 21(2)²= 84
The slopes at x – 2 & -2 are the same,
Hence the tangent will be parallel to each other.

Question 17. Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinates of the point.

Solution:

Given curve: y = x³
On differentiating w.r.t. x, we get
dy/dx = 3x² -(1)
Now let us assume that the point is (x₁, y₁)
dy/dx = 3x₁²
Also, slope of tangent at (x1, y1) is equal to y₁.
So, 3x₁²= y₁ -(2)
Also, (x1, y1) lies on y = x³ x3,so
y₁= x₁³ -(3)
From eq(2) & (3)
3x₁²= x₁³
3x₁²= x₁³= 0
x₁²(3 – x₁) = 0
For x₁= 0, y₁= x₁³ = (0)3 = 0
One such point is (0, 0)
For x₁= 3, y_1.= (3)3 = 27
Second point is (3, 27)

Question 18. For the curve y = 4x³– 2x⁵, find all the points at which the tangent passes through the origin.

Solution:

Given curve: y = 4x³– 3x⁵
Clearly at x = 0, y = 0, i.e the curve passes through origin.
Now the tangent also passes through origin.
Equation of a line passing through origin is y = mx.
Now tangent is touching the curve, so
y = mx will satisfy in curve.
mx = 4x³– 2x⁵ -(1)
Now dy/dx = 12x²– 10x⁴
Also m is the slope of tangent, so
m = 12x²– 10x⁴ -(2)
From eq(1) & (2),
(12x²– 10x⁴)x = 4x³– 3x⁵
x³(12 – 10x²) = x³(4 – 2x²) -(3)
For the first point, x = 0
For x₁= 0, y₁= 4x₁³ – 2x₁⁵ = 0
So, (0, 0) is one such point
Now for other roots of 3
12 – 10x²= 4 – 2x²
8 = 8x²
x²= 1
x = ±1
For x₂= 1, y₂ = 4x₂³ – 2x₂⁵ = 4(1)³ – 2(1)⁵ = 2
For x₃= -1, y₃ = 4x₃³ – 2x₃⁵ = 4(-1)³ – 2(-1)⁵ = -2
The other points are(1, 2) & (-1, -2)

Question 19. Find the points on the curve x²+ y²– 2x – 3 = 0 at which the tangents are parallel to the x-axis.

Solution:

Given curve: x²+ y²– 2x – 3 = 0
On differentiating w.r.t. x, we get
3x + 2y(dy/dx) – 2 – 0 = 0
$x+y\frac{dy}{dx}=1 \frac{dy}{dx}=\frac{1-x}{y}$
Given that the tangent are parallel to x-axis,
So, dy/dx = slope = 0
1 – x/y = 0
For x₁= 1,
x₁² + y₁² – 2x₁ – 3 = 0
(1)² + y₁² – 2(1) – 3 = 0
y₁²= 4
y₁²= ≠2
The points are (1, 2) & (1, -2)

Question 20. Find the equation of the normal at the point (am², am³) for the curve ay²= x³.

Solution:

Given curve: ay²= x³
On differentiating w.r.t. x, we get
2ay.dy/dx = 3x²
$\frac{dy}{dx}=\frac{3x^2}{2ay}$ & $\frac{-dx}{dy}=\frac{-2ay}{3x^2}$
$\frac{-dx}{dy}=\frac{-2a.am^3}{3.(am^2)^2}=\frac{-2}{3m}$
So, by point slope form, equation of normal is,
$y-y_1=\frac{-dx}{dy}(x-x_1)$
$y-am^3=\frac{-2}{3m}(x-am^2)$
3my – 3am⁴ = -2x + 2am²
Hence, 3my + 2x = 2am² + 3am⁴is the required equation of normal.

Question 21. Find the equation of the normals to the curve y = x³+ 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Solution:

Given curve: y = x³+ 2x + 6
On differentiating w.r.t. x, we get
$\frac{dy}{dx}=3x^2+2$
$\frac{-dx}{dy}=\frac{-1}{3x^2+2}$ -(Slope of normal)
Now, the normal are parallel to x + 14y + 4 = 0
$\frac{-1}{3x^2+2}=\frac{-1}{14}$
13 = 3x²+ 2
3x²= 12 ⇒ x²= 4
x = ±2
x₁= 2 & x₂= -2
For x₁= 2; y₁= $x_1^3+2x_1+6$ = (2)³+ 2(2) + 6 = 18
For x₂= -2; y₂= $x_2^2+2x_2+6$ = (-2)³+ 2(-2) + 6 = -6
Normal through (2,18) is
y – 18 = $\frac{-1}{14}(x-2)$
14y – 252 = -x + 2
14y + x = 254 is one such equation.
Normal through (-2, -6) is
y + 6 = $\frac{-1}{14}(x+2)$
14y + 84 = -x – 2
14y + x + 86 = 0 is the other equation of normal.

Question 22. Find the equations of tangent and normal to the parabola y²= 4ax at the point (at², 2at).

Solution:

Given parabola: y²= 4ax
On differentiating w.r.t. x, we get
$2y.\frac{dy}{dx}=4a$
$\frac{dy}{dx}=\frac{2a}{y} ;\frac{-dx}{dy}=\frac{-y}{2a}$
$\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t}$
Now, by point slope form, equation of tangent is,
y – y₁= $\frac{dy}{dx}(x-x_1)$
$y-2at=\frac{1}{t}(x-at^2)$
ty = x + at² is the equation of tangent to the parabola y²= 4ax at (at², 2at)
Now $\frac{-dx}{dy}=\frac{-2at}{2a}=-t$
Now, by point slope form, equation of normal is,
y – y₁= $\frac{-dx}{dy}(x-x_1)$
y – 2at = -t(x – at²)
y + xt = 2at + at³ is the equation of normal to the parabola y²= 4ax at (at², 2at)

Question 23. Prove that the curves x = y² and xy = k cut at right angles if 8k²= 1.

Solution:

Given curves: x = y²& xy = k
Two curves intersect at right angles f the tangents through their point
intersection is perpendicular to each other.
Now if tangents are perpendicular their product of their slopes will be equal to -1.
Curve 1: x = y²
1 = 2y.dy/dx
dy/dx = 1/2 y = m₁ -(1)
Curve 2: xy = k
y = k/x
$\frac{-k}{x^2}=m_2$
Let’s find their point of intersection
x = y²& xy = k
k/y = y²
y = k^1/3
x = y²
x = k^2/3
The point is (k^2/3, k^1/3)
m₁= 1/2k^1/3
For curves to be intersecting each other at right angles,
m₁m₂= -1
$\frac{1}{2k^\frac{1}{3}}.\frac{-1}{k^\frac{1}{3}}=1$
$2k^\frac{2}{3}=1$
8k²= 1
Hence proved

Question 24. Find the equations of the tangent and normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point (x_o, y_o).

Solution:

Given curve: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
On differentiating both sides with respect to x,
$\frac{2x}{a^2}-\frac{2y}{b^2}.\frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{xb^2}{ya^2}\&\frac{-dx}{dy}=\frac{-ya^2}{xb^2}$
Now, $\frac{dy}{dx}=\frac{x_ob^2}{y_0a^2}=m_T$
Equation of tangent by point slope form is,
$y-y_1=m_T(x-x_1)$
$y-y_0=\frac{x_0b^2}{y_0a^2}(x-x_0)$
$\frac{yy_0-y_0^2}{b^2}=\frac{xx_0-x_0^2}{a^2}$
$\frac{yy_0}{b^2}-\frac{xx_0}{a^2}=\frac{y_0^2}{a^2}=-1$ -((x₀,y₀) lie on $\frac{x^2}{a^2}-\frac{b^2}{b^2}=1$ )
$\frac{xx_0}{a^2}-\frac{yy_0}{b^2}=1$ is the equation of tangent.
Now, $\frac{-dx}{dy}=\frac{-y_0a^2}{x_0b^2}=m_N$
Equation of normal by point slope form is,
$y-y_1=m_N(x-x_1)$
$y-y_0=\frac{-y_0a^2}{x_0b^2}(x-x_0)$
x₀b²y – x₀b²y₀= -y₀a²x + y₀a²x₀
x₀b²y + y₀a²x = x₀y₀(a²+ b²) is the equation of normal

Question 25. Find the equation of the tangent to the curve y = $\sqrt{3x-2}$ which is parallel to the line 4x – 2y + 5 = 0.

Solution:

Given curve: $y=\sqrt{3x-2}$
On differentiating w.r.t. x, we get
$\frac{dy}{dx}=\frac{1}{2\sqrt{3x-2}}.3$
Now, it is given that the tangent is parallel to the line 4x – 2y + 5 = 0,
so their slopes must be equal.
Slope of 2y = 4x + 5 is 2.
So,
$\frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}=2$
$\frac{3}{4}=\sqrt{3x-2}$
9/16 = 3x – 2
x₁ = 41/48
Now, $y_1=\sqrt{3x_1-2}$
$y_1=\sqrt{3.\frac{41}{48}-2}=\frac{3}{4}$
The point is (41/48, 3/4)
Now by point slope form, the equation of tangent will be
y – y₁ = m(x – x₁)
$y-\frac{3}{4}=2(x-\frac{41}{48})$
$y-\frac{3}{4}=2x-\frac{41}{24}$
24y – 48x + 23 = 0 is the required equation of tangent.

Question 26. The slope of the normal to the curve y = 2x²+ 3 sin x at x = 0 is

(A) 3 (B) 1/3 (C)-3 (D) -1/3

Solution:

Given curve: y = 2x²+ 3 sin x
On differentiating w.r.t. x, we get
dy/dx = 4x + 3cos x
$\frac{-dx}{dy}=\frac{-1}{4x+3\cos x}$ -(Slope of normal)
$\frac{-dx}{dy}=\frac{-1}{4(0)+3\cos 0}=\frac{-1}{3}$
Hence, the correct option is D.

Question 27. The line y = x + 1 is a tangent to the curve y²= 4x at the point

(A) (1, 2) (B) (2, 1) (C) (1, -2) (D) (-1, 2)

Solution:

Given curve: y² = 4x
On differentiating w.r.t. x, we get
$2y\frac{dy}{dx}=4$
$\frac{dy}{dx}=\frac{2}{y}$
y = x + 1 is tangent, slope is 1, so, 2/y = 2
y₁= 2,
y₁² = 4x₁
x₁ = $\frac{2^2}{4}$
x₁= 1
So, (1, 2) is the point.
Hence, the correct option is A.

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Last Updated : 06 Apr, 2021

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

Question 15. Find the equation of the tangent line to the curve y = x²– 2x + 7 which is

Question 16. Show that the tangent to the curve y = 7x³+ 11 at the points where x = 2 and x = -2 are parallel.

Question 17. Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinates of the point.

Question 18. For the curve y = 4x³– 2x⁵, find all the points at which the tangent passes through the origin.

Question 19. Find the points on the curve x²+ y²– 2x – 3 = 0 at which the tangents are parallel to the x-axis.

Question 20. Find the equation of the normal at the point (am², am³) for the curve ay²= x³.

Question 21. Find the equation of the normals to the curve y = x³+ 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Question 22. Find the equations of tangent and normal to the parabola y²= 4ax at the point (at², 2at).

Question 23. Prove that the curves x = y² and xy = k cut at right angles if 8k²= 1.

Question 24. Find the equations of the tangent and normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point (x_o, y_o).

Question 25. Find the equation of the tangent to the curve y = $\sqrt{3x-2}$ which is parallel to the line 4x – 2y + 5 = 0.

Question 26. The slope of the normal to the curve y = 2x²+ 3 sin x at x = 0 is

(A) 3 (B) 1/3 (C)-3 (D) -1/3

Question 27. The line y = x + 1 is a tangent to the curve y²= 4x at the point

(A) (1, 2) (B) (2, 1) (C) (1, -2) (D) (-1, 2)

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

Question 15. Find the equation of the tangent line to the curve y = x2 – 2x + 7 which is

Question 16. Show that the tangent to the curve y = 7x3 + 11 at the points where x = 2 and x = -2 are parallel.

Question 17. Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinates of the point.

Question 18. For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.

Question 19. Find the points on the curve x2 + y2 – 2x – 3 = 0 at which the tangents are parallel to the x-axis.

Question 20. Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3.

Question 21. Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Question 22. Find the equations of tangent and normal to the parabola y2 = 4ax at the point (at2, 2at).

Question 23. Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1.

Question 24. Find the equations of the tangent and normal to the hyperbola at the point (xo, yo).

Question 25. Find the equation of the tangent to the curve y = which is parallel to the line 4x – 2y + 5 = 0.

Question 26. The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is

(A) 3 (B) 1/3 (C)-3 (D) -1/3

Question 27. The line y = x + 1 is a tangent to the curve y2 = 4x at the point

(A) (1, 2) (B) (2, 1) (C) (1, -2) (D) (-1, 2)

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Question 15. Find the equation of the tangent line to the curve y = x²– 2x + 7 which is

Question 16. Show that the tangent to the curve y = 7x³+ 11 at the points where x = 2 and x = -2 are parallel.

Question 17. Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinates of the point.

Question 18. For the curve y = 4x³– 2x⁵, find all the points at which the tangent passes through the origin.

Question 19. Find the points on the curve x²+ y²– 2x – 3 = 0 at which the tangents are parallel to the x-axis.

Question 20. Find the equation of the normal at the point (am², am³) for the curve ay²= x³.

Question 21. Find the equation of the normals to the curve y = x³+ 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Question 22. Find the equations of tangent and normal to the parabola y²= 4ax at the point (at², 2at).

Question 23. Prove that the curves x = y² and xy = k cut at right angles if 8k²= 1.

Question 24. Find the equations of the tangent and normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at the point (x_o, y_o).

Question 25. Find the equation of the tangent to the curve y = $\sqrt{3x-2}$ which is parallel to the line 4x – 2y + 5 = 0.

Question 26. The slope of the normal to the curve y = 2x²+ 3 sin x at x = 0 is

Question 27. The line y = x + 1 is a tangent to the curve y²= 4x at the point