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

### Chapter 6 Application of Derivatives -Exercise 6.3 | Set 1

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

**(i) y = x ^{4}– 6x^{3 }+ 13x^{2 }– 10x+ 5 at (0, 5)**

**(ii) y = x ^{4 }– 6x^{3 }+ 13x^{2 }– 10x + 5 at (1, 3)**

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

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

**(v) x = cos t, y = sin t at t = Ï€/4**

**Solution:**

(i)Given curvey = x

^{4 }– 6x^{3 }+ 13x^{2 }– 10x + 5Given point, (0, 5)

dy/dx = 4x

^{3 }– 18x^{2 }+ 26x – 10dy/dx = 4(0)

^{3 }– 18(0)^{2 }+ 26(0) – 10dy/dx = -10,

-dx/dy = 1/10

Now, with the help of points slope form

y – y

_{1 }= m(x – x_{1})y – 5 = -10(x – 0)

y + 10x = 5 is the required equation of the tangent

For equation of normal,

y – y1 =

y – 5 =

10y – x – 50 is the equation of normal.

(ii)Given curve: y = x^{4 }– 6x^{3 }+ 13x^{2 }– 10x + 5Given point is (1, 3)

dy/dx = 4x

^{3 }– 18x^{2 }+ 26x – 10dy/dx = 4(1)

^{3 }– 18(1)^{2 }+ 26(1) – 10dy/dx = 4 – 18 + 26 – 10 = 2

dy/dx = 2

-dx/dy = -1/2

Using point slope form, equation of tangent is

y – y

_{1 }= 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 – y

_{1 }= -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 = x^{3}Given point is (1, 1)

dy/dx = 3x

^{2}dy/dx = 3(1)

^{2 }= 3dy/dx = 3 & -dx/dy = -1/3

Using point slope form, equation of tangent is y – y

_{1 }= dy/dx(x – x_{1})y – y

_{1 }= dy/dx(x – x_{1})y – 1 = 3(x – 1)

y – 3x + 2 = 0 is the equation of tangent

Using point slope form, equation of normal is

y – y1 =

y – 1 =

3y – 3 = -x + 1

3y + x = 4 is the equation of normal.

(iv)Given curve: y = x^{2}Given point (0, 0)

dy/dx = 2x

dy/dx = 0

dy/dx = 0 & -dx/dy = not defined is

y – y

_{1 }= dy/dx(x – x_{1})y – 0 = 0(x – 0)

y = 0 is the equation of tangent.

Using point slope form, equation of normal is

y – y

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

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 tPoint t = Ï€/4

-(1)

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

dy/dx = -1 & -dx/dy = 1

Now for t = Ï€/4,

y

_{1 }= sin t = sin(Ï€/4) = 1/âˆš2x

_{1 }= cos t = cos(Ï€/4) = 1/âˆš2The point is (1/âˆš2, 1/âˆš2)

y – y

_{1 }=y – (1/âˆš2) = -1(x – 1/âˆš2)

y – 1/âˆš2 = -x + 1/âˆš2

x + y = âˆš2 is the equation of normal is

y – 1/âˆš2 = 1(x – 1/âˆš2)

x = y is the equation of normal.

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

**(i) Parallel to line 2x – y + 9 = 0 **

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

**Solution:**

Given curve: y = x

^{2 }– 2x + 7On 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

x

_{1 }= 2Corresponding to x

_{1 }= 2,y

_{1}= x_{1}^{2}– 2x_{1}+ 7y

_{1 }= (2)^{2 }– 2(2) + 7y

_{1 }= 7The point of contact is (2, 7).

Using point slope form, equation of tangent is

y – y

_{1 }=y – 7 = 2(x – 2)

y – 2x = 3 is the equation of tangent.

(ii)Tangent is perpendicular to the line 5y – 15x = 13That 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 = x

^{2 }– 2x + 7dy/dx = 2x – 2

On comparing dy/dx with slope, we get

2x – 2 = -1/3

6x – 6 = -1

6x = 5

x

_{1}= 5/6For x

_{1 }= 5/6,y

_{1 }= x_{1}^{2}– 2x_{1}+ 7y

_{1 }= (5/6)_{1}^{2}– 2(5/6)_{1}+ 7y

_{1 }= 217/36Now using point slope form, equation of tangent is

y – y

_{1}= m(x – x_{1})36y – 217 = -12x + 10

36y + 12x = 227 is the required equation of tangent.

### Question 16. Show that the tangent to the curve y = 7x^{3 }+ 11 at the points where x = 2 and x = -2 are parallel.

**Solution:**

Given curve: y = 7x

^{3 }+ 11On differentiating w.r.t. x, we get

dy/dx = 21x

^{2}dy/dx = 21(2)

^{2 }= 84The 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 = x^{3} at which the slope of the tangent is equal to the y-coordinates of the point.

**Solution:**

Given curve: y = x

^{3}On differentiating w.r.t. x, we get

dy/dx = 3x

^{2}-(1)Now let us assume that the point is (x

_{1}, y_{1})dy/dx = 3x

_{1}^{2}Also, slope of tangent at (x1, y1) is equal to y

_{1}.So, 3x

_{1}^{2 }= y_{1}-(2)Also, (x1, y1) lies on y = x

^{3}x3,soy

_{1 }= x_{1}^{3}-(3)From eq(2) & (3)

3x

_{1}^{2 }= x_{1}^{3}3x

_{1}^{2 }= x_{1}^{3 }= 0x

_{1}^{2}(3 – x_{1}) = 0For x

_{1 }= 0, y_{1 }= x_{1}^{3}= (0)3 = 0One such point is (0, 0)

For x

_{1 }= 3, y_{1. }= (3)3 = 27Second point is (3, 27)

### Question 18. For the curve y = 4x^{3 }– 2x^{5}, find all the points at which the tangent passes through the origin.

**Solution:**

Given curve: y = 4x

^{3 }– 3x^{5}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 = 4x

^{3 }– 2x^{5}-(1)Now dy/dx = 12x

^{2 }– 10x^{4}Also m is the slope of tangent, so

m = 12x

^{2 }– 10x^{4}-(2)From eq(1) & (2),

(12x

^{2 }– 10x^{4})x = 4x^{3 }– 3x^{5}x

^{3}(12 – 10x^{2}) = x^{3}(4 – 2x^{2}) -(3)For the first point, x = 0

For x

_{1 }= 0, y_{1 }= 4x_{1}^{3}– 2x_{1}^{5}= 0So, (0, 0) is one such point

Now for other roots of 3

12 – 10x

^{2 }= 4 – 2x^{2}8 = 8x

^{2}x

^{2 }= 1x = Â±1

For x

_{2 }= 1, y_{2}= 4x_{2}^{3}– 2x_{2}^{5}= 4(1)^{3}– 2(1)^{5}= 2For x

_{3 }= -1, y_{3}= 4x_{3}^{3}– 2x_{3}^{5}= 4(-1)^{3}– 2(-1)^{5}= -2The other points are(1, 2) & (-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.

**Solution:**

Given curve: x

^{2 }+ y^{2 }– 2x – 3 = 0On differentiating w.r.t. x, we get

3x + 2y(dy/dx) – 2 – 0 = 0

Given that the tangent are parallel to x-axis,

So, dy/dx = slope = 0

1 – x/y = 0

For x

_{1 }= 1,x

_{1}^{2}+ y_{1}^{2}– 2x_{1}– 3 = 0(1)

^{2}+ y_{1}^{2}– 2(1) – 3 = 0y

_{1}^{2 }= 4y

_{1}^{2 }= â‰ 2The points are (1, 2) & (1, -2)

### Question 20. Find the equation of the normal at the point (am^{2}, am^{3}) for the curve ay^{2 }= x^{3}.

**Solution:**

Given curve: ay

^{2 }= x^{3}On differentiating w.r.t. x, we get

2ay.dy/dx = 3x

^{2}&

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

3my – 3am

^{4}= -2x + 2am^{2}Hence, 3my + 2x = 2am

^{2}+ 3am^{4 }is the required equation of normal.

### 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.

**Solution:**

Given curve: y = x

^{3 }+ 2x + 6On differentiating w.r.t. x, we get

-(Slope of normal)

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

13 = 3x

^{2 }+ 23x

^{2 }= 12 â‡’ x^{2 }= 4x = Â±2

x

_{1 }= 2 & x_{2 }= -2For x

_{1 }= 2; y_{1 }= = (2)^{3 }+ 2(2) + 6 = 18For x

_{2 }= -2; y_{2 }= = (-2)^{3 }+ 2(-2) + 6 = -6Normal through (2,18) is

y – 18 =

14y – 252 = -x + 2

14y + x = 254 is one such equation.

Normal through (-2, -6) is

y + 6 =

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 y^{2 }= 4ax at the point (at^{2}, 2at).

**Solution:**

Given parabola: y

^{2 }= 4axOn differentiating w.r.t. x, we get

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

y – y

_{1 }=ty = x + at

^{2}is the equation of tangent to the parabola y^{2 }= 4ax at (at^{2}, 2at)Now

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

y – y

_{1 }=y – 2at = -t(x – at

^{2})y + xt = 2at + at

^{3}is the equation of normal to the parabola y^{2 }= 4ax at (at^{2}, 2at)

### Question 23. Prove that the curves x = y^{2} and xy = k cut at right angles if 8k^{2 }= 1.

**Solution:**

Given curves: x = y

^{2 }& xy = kTwo 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 = y

^{2}1 = 2y.dy/dx

dy/dx = 1/2 y = m

_{1}-(1)Curve 2: xy = k

y = k/x

Let’s find their point of intersection

x = y

^{2 }& xy = kk/y = y

^{2}y = k

^{1/3}x = y

^{2}x = k

^{2/3}The point is (k

^{2/3}, k^{1/3})m

_{1 }= 1/2k^{1/3}For curves to be intersecting each other at right angles,

m

_{1}m_{2 }= -18k

^{2 }= 1Hence proved

### Question 24. Find the equations of the tangent and normal to the hyperbola at the point (x_{o}, y_{o}).

**Solution:**

Given curve:

On differentiating both sides with respect to x,

Now,

Equation of tangent by point slope form is,

-((x

_{0},y_{0}) lie on )is the equation of tangent.

Now,

Equation of normal by point slope form is,

x

_{0}b^{2}y – x_{0}b^{2}y_{0 }= -y_{0}a^{2}x + y_{0}a^{2}x_{0}x

_{0}b^{2}y + y_{0}a^{2}x = x_{0}y_{0}(a^{2 }+ b^{2}) is the equation of normal

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

**Solution:**

Given curve:

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

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,

9/16 = 3x – 2

x

_{1}= 41/48Now,

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

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

y – y

_{1}= m(x – x_{1})24y – 48x + 23 = 0 is the required equation of tangent.

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

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

**Solution:**

Given curve: y = 2x

^{2 }+ 3 sin xOn differentiating w.r.t. x, we get

dy/dx = 4x + 3cos x

-(Slope of normal)

Hence, the correct option is D.

### Question 27. The line y = x + 1 is a tangent to the curve y^{2 }= 4x at the point

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

**Solution:**

Given curve: y

^{2}= 4xOn differentiating w.r.t. x, we get

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

y

_{1 }= 2,y

_{1}^{2}= 4x_{1}x

_{1}=x

_{1 }= 1So, (1, 2) is the point.

Hence, the correct option is A.

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