NCERT Solutions Class 12 maths Chapter-6 (Applications of Derivatives)Exercise 6.3

Jolly September 06, 2022

NCERT Solutions Class 12 maths Chapter-6 (Applications of Derivatives)Exercise 6.3

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-6 (Applications of Derivatives)Exercise 6.3 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Solutions Class 12 maths Chapter-6 (Applications of Derivatives)Exercise 6.3

Exercise 6.3 

Set 1

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 ,y = \frac{x-1}{x-2}, x  ≠ 2 at x = 10.

Solution:

Given curve: y=\frac{x-1}{x-2}

y=\frac{x-2+1}{x-2}=1+\frac{1}{x-2}

On differentiating w.r.t x, we get

\frac{dy}{dx}=0-\frac{1}{(x-2)^2}

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

\frac{dy}{dx}=\frac{-1}{(10-2)^2}=\frac{-1}{64}

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

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

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

\frac{dy}{dx}=3(2)^2-1=11

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 \frac{-dx}{dy}

\frac{-dx}{dy}=-\frac{\frac{dx}{dθ}}{\frac{dy}{dθ}}    -(1)

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

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

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

\frac{-dx}{dy}=\frac{\cosθ}{\sinθ}=\cotθ

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

\frac{-dx}{dy}=\cot(\frac{π}{4})=1

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 \frac{-dx}{dy}

\frac{-dx}{dy}=\frac{\frac{-dx}{dθ}}{\frac{dy}{dθ}}          -(1)

\frac{dx}{dθ}=\frac{d(1-a\sinθ)}{dθ}=-a\cosθ

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

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

\frac{-dx}{dy}=\frac{-a}{2b\sinθ}

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

\frac{-dx}{dy}=\frac{-a}{2b\sin\frac{π}{2}}=\frac{-a}{2b}

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 = \frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{4-2}=\frac{4}{2}=2

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  = \frac{1}{x-1}[Tex]  [/Tex], x ≠ 1.

Solution:

Given curve: y = \frac{1}{x-1}

\frac{dy}{dx}=\frac{-1}{(x-1)^2}                  -(1)

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

dy/dx = -1                  -(1)

From 1 & 2, we get

-1 = \frac{-1}{(x-1)^2}

(x – 1)2 = 1

x = 1 ± 1

x1 = 2 & x2 = 0

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

y_1=\frac{1}{x-1}=\frac{1}{2-1}=1

y_2=\frac{1}{x_2-1}=\frac{1}{0-1}=-1

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 ,y = \frac{1}{x-3}, 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 y=\frac{1}{x^2-2x+3}. .

Solution:

Given curve,

y=\frac{1}{x^2-2x+3} 

On differentiating w.r.t x, we get

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

Given slope = 0 = dy/dx  

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

x – 1 = 0

x = 1

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

So the equation of tangent from point slope from is 

y – y1 = m(x – x1)

y-\frac{1}{2}= 0(x – 1)

2y – 1 = 0

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

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

Solution:

Given curve:

\frac{x^2}{9}+\frac{y^2}{16}=1         -(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,

\frac{2x}{9}+\frac{2y}{16}.\frac{dy}{dx}=0 

\frac{dy}{dx}=\frac{-16x}{9y}

Now slope = 0, so

\frac{-16x}{9y}=0

= x1 = 0

For x1 = 0,   

\frac{0^2}{9}+\frac{y_1^2}{16}=1

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

\frac{2x}{9}.\frac{dx}{dy}+\frac{2y}{16}=0 

\frac{dx}{dy}=0,\frac{-9y}{16x}=0

y2 = 0

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

x22 = 9

x2 = ±3

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

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, y1. = (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 + 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

\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 = k1/3

x = y2      

x = k2/3

The point is (k2/3, k1/3)

m1 = 1/2k1/3

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 (xo, yo).

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.

Chapter-6 (Applications of Derivatives)

Tags: