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

Question 1. Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4.

Solution:

Given curve: y = 3x⁴ – 4x

On differentiating w.r.t x, we get

dy/dx = 12x³ – 4

Now, we find the slope of the tangent to the given curve at x = 4 is

= dy/dx = 12(4)³ – 4 = 764

Hence, the slope is 764

Question 2. Find the slope of the tangent to the curve [Tex]y = \frac{x-1}{x-2} [/Tex], x ≠ 2 at x = 10.

Solution:

Given curve: [Tex]y=\frac{x-1}{x-2}[/Tex]

[Tex]y=\frac{x-2+1}{x-2}=1+\frac{1}{x-2}[/Tex]

On differentiating w.r.t x, we get

[Tex]\frac{dy}{dx}=0-\frac{1}{(x-2)^2}[/Tex]

Now, we find the slope of the tangent to the given curve at x = 10 is

[Tex]\frac{dy}{dx}=\frac{-1}{(10-2)^2}=\frac{-1}{64}[/Tex]

Hence, the slope is -1/64

Question 3. Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x-coordinate is 2. 

Solution:

Given curve: y = x³ – x + 1

On differentiating w.r.t x, we get

[Tex]\frac{dy}{dx}=\frac{d(x^3-x+1)}{dx}=3x^2-1[/Tex]

Now, we find the slope of the tangent to the given curve at x = 2 is

[Tex]\frac{dy}{dx}=3(2)^2-1=11[/Tex]

Hence, the slope is 11

Question 4. Find the slope of the tangent to the curve y = x³ –3x + 2 at the point whose x-coordinate is 3.

Solution:

Given curve: y = x³ – 3x + 2

On differentiating w.r.t x, we get

dy/dx = 3x² – 3

Now, we find the slope of the tangent to the given curve at x = 3 is

dy/dx = 3(3)² – 3 = 24

Hence, the slope is 24

Question 5. Find the slope of the normal to the curve x = acos³θ, y = asin³θ at θ = π/4.

Solution:

Given curve: x = acos³θ = f(θ)

y = asin³θ = g(θ)

To find slope of the normal of the curve at θ = π/4

Now, slope of the normal is [Tex]\frac{-dx}{dy}[/Tex]

[Tex]\frac{-dx}{dy}=-\frac{\frac{dx}{dθ}}{\frac{dy}{dθ}}   [/Tex]    -(1)

[Tex]\frac{dx}{d\theta} =\frac{d(a\cos^3θ) }{dθ} [/Tex] = a.3.cos² θ.(-sin θ)

[Tex]\frac{dy}{dθ}=\frac{d(a\sin^3θ)}{dθ} [/Tex] = a.3sin² θ.cos θ

[Tex]\frac{-dx}{dθ}=\frac{a.3.\cos^2θ.(-\sinθ)}{a.3.\sin^2θ.\cosθ}   [/Tex]      -(using eq(1))

[Tex]\frac{-dx}{dy}=\frac{\cosθ}{\sinθ}=\cotθ[/Tex]

Now, we find the slope of the tangent to the given curve at θ = π/4 is

[Tex]\frac{-dx}{dy}=\cot(\frac{π}{4})=1[/Tex]

The slope of normal of the parametric curve

Hence, the slope is 1

Question 6. Find the slope of the normal to the curve x = 1 – asinθ, y = bcos²θ at θ = π/2.

Solution:

Given curve: x = 1 – a sinθ

y = b cos2θ

Now, slope of normal is [Tex]\frac{-dx}{dy}[/Tex]

[Tex]\frac{-dx}{dy}=\frac{\frac{-dx}{dθ}}{\frac{dy}{dθ}}  [/Tex]          -(1)

[Tex]\frac{dx}{dθ}=\frac{d(1-a\sinθ)}{dθ}=-a\cosθ[/Tex]

[Tex]\frac{dy}{dθ}=\frac{d(b\cos^2θ)}{dθ}=-2b\cosθ\sinθ[/Tex]

[Tex]\frac{-dx}{dy}=\frac{-a\cosθ}{-2b\cosθ\sinθ}  [/Tex]        -(using eq(1))

[Tex]\frac{-dx}{dy}=\frac{-a}{2b\sinθ}[/Tex]

Now, we find the slope of the tangent to the given curve at θ = π/2 is

[Tex]\frac{-dx}{dy}=\frac{-a}{2b\sin\frac{π}{2}}=\frac{-a}{2b}[/Tex]

Hence, the slope is -a/2b.

Question 7. Find points at which the tangent to the curve y = x³ – 3x² – 9x + 7 is parallel to the x-axis.

Solution:

Given curve: y = x³ – 3x² – 9x + 7

On differentiating w.r.t x, we get

dy/dx = 3x² – 6x – 9

For tangent to be parallel to x-axis, slope is 0. So dy/dx = 0.

3x² – 6x – 9 = 0

3(x² – 2x – 3) = 0

3(x² + x – 3x – 3) = 0

3(x(x + 1) – 3(x + 1)) = 0

3(x + 1)(x – 3) = 0

x = -1 or x = 3

For x = -1, y = (-1)³ – 3(-1)² – 9(-1) + 7

x = -1, y = -1 – 3 + 9 + 7 = 12

Hence, the first point is (-1, 12)

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).

Solution:

Given curve: y = (x – 2)²

On differentiating w.r.t x, we get

dy/dx = 2(x – 2)                 -(1)

Given that, the tangent is parallel to the chord joining the points (2, 0) & (4, 4)

Slope of the chord = [Tex]\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{4-2}=\frac{4}{2}=2[/Tex]

Now equality dy/dx = slope of chord

2(x – 2) = 2

x – 2 = 1

x = 3

y = (x – 2)²

y = (3 – 2)² = 1

Hence, the point on the curve y = (x – 2)² is (3, 1)

Question 9. Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11. 

Solution:

Given curve: y = x³ – 11x + 5

Given tangent: y = x – 11

From the given tangent, we can find out the slope comparing y = x – 11 with y = mx + c, we get

Slope(m) = 1

Now y = x³ – 11x + 5

dy/dx = 3x² – 11                  -(1)

dy/dx = slope = 1

So, from eq(1), we get  

3x² – 11 = 1

3x² = 12

x² = 4

x = ±2

If x = +2, y = 2³ – 11(2) + 5 = -9

If x = -2, y = (-2)³ – 11(-2) + 5 = 19

The points must lie on the tangent as well.

Only (2,-9) is satisfying the tangent equation.

So the point on the curve whose tangent is y = x – 11 is (2,-9). 

Question 10. Find the equation of all lines having slope –1 that are tangents to the curve y = [Tex]\frac{1}{x-1} [/Tex][Tex]  [/Tex], x ≠ 1.

Solution:

Given curve: y = [Tex]\frac{1}{x-1}[/Tex]

[Tex]\frac{dy}{dx}=\frac{-1}{(x-1)^2}  [/Tex]                  -(1)

Now given slope = -1 & we know that dy/dx = slope, so

dy/dx = -1                  -(1)

From 1 & 2, we get

-1 = [Tex]\frac{-1}{(x-1)^2}[/Tex]

(x – 1)² = 1

x = 1 ± 1

x₁ = 2 & x₂ = 0

Now corresponding to these x₁ & x₂ we need to find out y₁ & y₂ 

[Tex]y_1=\frac{1}{x-1}=\frac{1}{2-1}=1[/Tex]

[Tex]y_2=\frac{1}{x_2-1}=\frac{1}{0-1}=-1[/Tex]

The points are (2, 1) & (0, -1)

Now equations slope is -1

Using point slope form the first tangent equation is 

(y – y₁) = m(x – x₁)

y – 1 = -1(x – 2)

= x + y = 3

Using point slope from the second tangent equation is 

(y – y₂) = m(x – x₂)

y – (-1) = -1(x – 0)

= x + y + 1 = 0 

Question 11. Find the equation of all lines having slope 2 which are tangents to the curve [Tex]y = \frac{1}{x-3} [/Tex], x ≠ 3.

Solution:

Given curve: y = 1/(x – 3) 

dy/dx = -1/(x – 3)² = slope               -(dy/dx is slope)

Now given slope is 2, so

dy/dx = -1/(x – 3)² = 2

(x – 3)² = -1/2              -(1) (not possible)

Now because there is no real value of x which can satisfy 1, therefore no such tangent exists on the curve y = 1/x – 3 whose is 2.

Question 12. Find the equations of lines having slope 0 which are tangent to the curve [Tex]y=\frac{1}{x^2-2x+3} [/Tex].

Solution:

Given curve,

[Tex]y=\frac{1}{x^2-2x+3}    [/Tex] 

On differentiating w.r.t x, we get

[Tex]\frac{dy}{dx}=\frac{-1}{(x^2-2x+3)}.(2x-2)    [/Tex]      -(chain rule)

Given slope = 0 = dy/dx  

So,[Tex]\frac{dy}{dx}=\frac{-2(x-1)}{(x^2-2x+3)^2}=0 [/Tex]

x – 1 = 0

x = 1

For x = 1, y = [Tex]\frac{1}{1^2-2(1)+3}=\frac{1}{2}[/Tex]

So the equation of tangent from point slope from is 

y – y₁ = m(x – x₁)

[Tex]y-\frac{1}{2}    [/Tex]= 0(x – 1)

2y – 1 = 0

Question 13. Find points on the curve [Tex]\frac{x^2}{9}+\frac{y^2}{16}=1 [/Tex] at which the tangents are 

(i) Parallel to x-axis            (ii)Parallel to y-axis

Solution:

Given curve:

[Tex]\frac{x^2}{9}+\frac{y^2}{16}=1   [/Tex]         -(1)

(i) If tangent is parallel to x-axis then it means slope is 0 or dy/dx = 0

On differentiating both sides of equations (1) we get,

[Tex]\frac{2x}{9}+\frac{2y}{16}.\frac{dy}{dx}=0   [/Tex] 

[Tex]\frac{dy}{dx}=\frac{-16x}{9y}[/Tex]

Now slope = 0, so

[Tex]\frac{-16x}{9y}=0[/Tex]

= x₁ = 0

For x₁ = 0,   

[Tex]\frac{0^2}{9}+\frac{y_1^2}{16}=1[/Tex]

y₁² = 16

y₁ = ±4

The coordinates are (0, 4) & (0, -4)

(ii) Now, if tangent is parallel to y-axis to the dy/dx or slope is not defined or dy/dx = 0

On differentiating equations(1) with respect to y, we get

[Tex]\frac{2x}{9}.\frac{dx}{dy}+\frac{2y}{16}=0   [/Tex] 

[Tex]\frac{dx}{dy}=0,\frac{-9y}{16x}=0[/Tex]

y₂ = 0

For y₂ = 0, [Tex]\frac{x_2^2}{9}+\frac{0}{16}=1[/Tex]

x₂² = 9

x₂ = ±3

Hence, the coordinates are (3, 0) & (-3, 0)

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 = [Tex]\frac{-dx}{dy}(x-x_1)[/Tex]

y – 5 = [Tex]\frac{1}{10}(x-0)[/Tex]

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 = [Tex]\frac{-dx}{dy}(x-x_1)[/Tex]

y – 1 = [Tex]\frac{-1}{3}(x-1)[/Tex]

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 = [Tex]\frac{-dx}{dy}(x-x_1)   [/Tex]           -(1)

-dx\dy is undefined, so we can write eq(1) as 

[Tex]\frac{-dy}{dx}(y-y_1)=x-x_1[/Tex]

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

[Tex]\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}      [/Tex]    -(1)

[Tex]\frac{dx}{dt}=\frac{d(\cos t)}{dt}=-\sin t[/Tex]

[Tex]\frac{dy}{dt}=\frac{d(\sin t)}{dt}=\cos t[/Tex]

On putting these values in eq(1), we get

[Tex]\frac{dy}{dx}=\frac{\cos t}{-\sin t}=-\cot t[/Tex]

[Tex]\frac{dy}{dx}=-\cot \frac{π}{4}=-1[/Tex]

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 = [Tex]\frac{dy}{dx}(x-x_1)[/Tex]

y – (1/√2) = -1(x – 1/√2)

y – 1/√2 = -x + 1/√2

x + y = √2 is the equation of normal is 

[Tex]y-y_1=\frac{-dx}{dy}(x-x_1)[/Tex]

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 = [Tex]\frac{dy}{dx}(x-x_1)[/Tex]

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)

[Tex]y-\frac{217}{36}=\frac{-1}{3}(x-\frac{5}{6})[/Tex]

[Tex]\frac{36y-217}{36}=-1\frac{(6x-5)}{6.3}[/Tex]

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

[Tex]x+y\frac{dy}{dx}=1 \frac{dy}{dx}=\frac{1-x}{y}[/Tex]

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

[Tex]\frac{dy}{dx}=\frac{3x^2}{2ay} [/Tex] & [Tex]\frac{-dx}{dy}=\frac{-2ay}{3x^2}[/Tex]

[Tex]\frac{-dx}{dy}=\frac{-2a.am^3}{3.(am^2)^2}=\frac{-2}{3m}[/Tex]

So, by point slope form, equation of normal is,

[Tex]y-y_1=\frac{-dx}{dy}(x-x_1)   [/Tex] 

[Tex]y-am^3=\frac{-2}{3m}(x-am^2)[/Tex]

3my – 3am⁴ = -2x + 2am²

Hence, 3my + 2x = 2am² + 3am⁴ is 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

[Tex]\frac{dy}{dx}=3x^2+2[/Tex]

[Tex]\frac{-dx}{dy}=\frac{-1}{3x^2+2}   [/Tex]       -(Slope of normal)

Now, the normal are parallel to x + 14y + 4 = 0

[Tex]\frac{-1}{3x^2+2}=\frac{-1}{14}[/Tex]

13 = 3x² + 2            

3x² = 12 ⇒  x² = 4

x = ±2

x₁ = 2 & x₂ = -2

For x₁ = 2; y₁ = [Tex]x_1^3+2x_1+6 [/Tex] = (2)³ + 2(2) + 6 = 18

For x₂ = -2; y₂ = [Tex]x_2^2+2x_2+6 [/Tex] = (-2)³ + 2(-2) + 6 = -6

Normal through (2,18) is 

y – 18 = [Tex]\frac{-1}{14}(x-2)[/Tex]

14y – 252 = -x + 2

14y + x = 254 is one such equation.

Normal through (-2, -6) is 

y + 6 = [Tex]\frac{-1}{14}(x+2)[/Tex]

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

[Tex]2y.\frac{dy}{dx}=4a[/Tex]

[Tex]\frac{dy}{dx}=\frac{2a}{y}  ;\frac{-dx}{dy}=\frac{-y}{2a}[/Tex]

[Tex]\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t}[/Tex]

Now, by point slope form, equation of tangent is,

y – y₁ = [Tex]\frac{dy}{dx}(x-x_1)[/Tex]

[Tex]y-2at=\frac{1}{t}(x-at^2)[/Tex]

ty = x + at² is the equation of tangent to the parabola y² = 4ax at (at², 2at)

Now [Tex]\frac{-dx}{dy}=\frac{-2at}{2a}=-t[/Tex]

Now, by point slope form, equation of normal is,

y – y₁ = [Tex]\frac{-dx}{dy}(x-x_1)[/Tex]

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

[Tex]\frac{-k}{x^2}=m_2[/Tex]

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

[Tex]\frac{1}{2k^\frac{1}{3}}.\frac{-1}{k^\frac{1}{3}}=1[/Tex]

[Tex]2k^\frac{2}{3}=1[/Tex]

8k² = 1  

Hence proved        

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

Solution:

Given curve: [Tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/Tex]

On differentiating both sides with respect to x,

[Tex]\frac{2x}{a^2}-\frac{2y}{b^2}.\frac{dy}{dx}=0   [/Tex] 

[Tex]\frac{dy}{dx}=\frac{xb^2}{ya^2}\&\frac{-dx}{dy}=\frac{-ya^2}{xb^2}[/Tex]

Now, [Tex]\frac{dy}{dx}=\frac{x_ob^2}{y_0a^2}=m_T[/Tex]

Equation of tangent by point slope form is,

[Tex]y-y_1=m_T(x-x_1)[/Tex]

[Tex]y-y_0=\frac{x_0b^2}{y_0a^2}(x-x_0)[/Tex]

[Tex]\frac{yy_0-y_0^2}{b^2}=\frac{xx_0-x_0^2}{a^2}[/Tex]

[Tex]\frac{yy_0}{b^2}-\frac{xx_0}{a^2}=\frac{y_0^2}{a^2}=-1   [/Tex]      -((x₀,y₀) lie on [Tex]\frac{x^2}{a^2}-\frac{b^2}{b^2}=1 [/Tex])

[Tex]\frac{xx_0}{a^2}-\frac{yy_0}{b^2}=1 [/Tex] is the equation of tangent.

Now, [Tex]\frac{-dx}{dy}=\frac{-y_0a^2}{x_0b^2}=m_N[/Tex]

Equation of normal by point slope form is,

[Tex]y-y_1=m_N(x-x_1)[/Tex]

[Tex]y-y_0=\frac{-y_0a^2}{x_0b^2}(x-x_0)[/Tex]

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 = [Tex]\sqrt{3x-2} [/Tex] which is parallel to the line 4x – 2y + 5 = 0.

Solution:

Given curve: [Tex]y=\sqrt{3x-2}[/Tex]

On differentiating w.r.t. x, we get

[Tex]\frac{dy}{dx}=\frac{1}{2\sqrt{3x-2}}.3[/Tex]

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,

[Tex]\frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}=2[/Tex]

[Tex]\frac{3}{4}=\sqrt{3x-2}[/Tex]

9/16 = 3x – 2

x₁ = 41/48

Now, [Tex]y_1=\sqrt{3x_1-2}[/Tex]

[Tex]y_1=\sqrt{3.\frac{41}{48}-2}=\frac{3}{4}[/Tex]

The point is (41/48, 3/4)

Now by point slope form, the equation of tangent will be 

y – y₁ = m(x – x₁)

[Tex]y-\frac{3}{4}=2(x-\frac{41}{48})[/Tex]

[Tex]y-\frac{3}{4}=2x-\frac{41}{24}[/Tex]

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

[Tex]\frac{-dx}{dy}=\frac{-1}{4x+3\cos x}   [/Tex]               -(Slope of normal)

[Tex]\frac{-dx}{dy}=\frac{-1}{4(0)+3\cos 0}=\frac{-1}{3}[/Tex]

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

[Tex]2y\frac{dy}{dx}=4 [/Tex]       

[Tex]\frac{dy}{dx}=\frac{2}{y}[/Tex]

y = x + 1 is tangent, slope is 1, so, 2/y = 2

y₁ = 2,

y₁² = 4x₁

x₁ = [Tex]\frac{2^2}{4}   [/Tex] 

x₁ = 1

So, (1, 2) is the point.

Hence, the correct option is A.