# Class 12 RD Sharma Solutions – Chapter 16 Tangents and Normals – Exercise 16.2 | Set 2

### Question 7. Find the equation of the normal to the curve ay^{2} = x^{3} at the point (am^{2}, am^{3}).

**Solution:**

We have,

ay

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

2aydy/dx = 3x

^{2}dy/dx = 3x

^{2}/2aySlope of tangent =

Given (x

_{1}, y_{1}) = (am^{2}, am^{3})The equation of normal is,

y – y

_{1}= -1/m (x – x_{1})y – a m

^{3}= -2m/3 (x – am^{2})3my – 3am

^{4}= – 2x + 2am^{2}2x + 3my – am

^{2}(2 + 3m^{2}) = 0

### Question 8. The equation of the tangent at (2, 3) on the curve y^{2} = ax^{3} + b is y = 4x âˆ’ 5. Find the values of a and b.

**Solution:**

We have,

y

^{2}= ax^{3}+ bOn differentiating both sides w.r.t. x, we get

2y dy/dx = 3ax

^{2}dy/dx = 3ax

^{2}/2ySlope of tangent, m =

The equation of tangent is given by y â€“ y

_{1}= m (tangent) (x â€“ x_{1})Now compare the slope of a tangent with the given equation

2a = 4

a = 2

Now (2, 3) lies on the curve, these points must satisfy

32 = 2 (23) + b

b = â€“ 7

### Question 9. Find the equation of the tangent line to the curve y = x^{2} + 4x âˆ’ 16 which is parallel to the line 3x âˆ’ y + 1 = 0.

**Solution:**

We have,

y = x

^{2}+ 4x âˆ’ 16Let (a, b) be the point of intersection of both the curve and the tangent.

Since (a, b) lies on curve, we get

b = a

^{2}+ 4a âˆ’ 16Now, x

^{2}+ 4x âˆ’ 16dy/dx = 2x + 4

Slope of tangent =

Given that the tangent is parallel to the line we have,

Slope of tangent = Slope of the given line

=> 2a + 4 = 3

=> 2 a = -1

=> a = -1/2

From eq(1), we get

b = 1/4 – 2 – 16 = -71/4

Now, slope of tangent, m = 3

(a, b) = (-1/2, -71/4)

The equation of tangent is,

y – y

_{1}= m (x – x_{1})y + 71/4 = 3 (x + 1/2)

4y + 71 = 12x + 6

12x – 4y – 65 = 0

### Question 10. Find an equation of normal line to the curve y = x^{3} + 2x + 6 which is parallel to the line x+ 14y + 4 = 0.

**Solution:**

We have,

y = x

^{3}+ 2x + 6Let (a, b) be a point on the curve where we need to find the normal.

Slope of the given line = -1/14

Since the point lies on the curve, we get

b = a

^{3}+ 2a + 6Now, y = x

^{3 }+ 2x + 6dy/dx = 3 x

^{2 }+ 2Slope of the tangent, m =

Slope of the normal =

Given that, slope of the normal = slope of the given line, we have

3a

^{2}+ 2 = 143a

^{2}= 12a

^{2}= 4a = Â±2

So, b = 18 or -6.

And slope of the normal = -1/14

When a = 2 and b = 18, we have

y – y

_{1 }= m (x – x_{1})y – 18 = -1/14 (x – 2)

14y – 252 = -x + 2

x + 14y – 254 = 0

When a = -2 and b = -6, we have

y – y

_{1}= m (x – x_{1})y + 6 = -1/14 (x + 2)

14y + 84 = -x – 2

x + 14y + 86 = 0

### Question 11. Determine the equation(s) of tangent (s) line to the curve y = 4x^{3} âˆ’ 3x + 5 which are perpendicular to the line 9y + x + 3 = 0.

**Solution:**

Let (a, b) be a point on the curve where we need to find the tangent(s).

Slope of the given line = -1/9

Since, tangent is perpendicular to the given line,

Slope of the tangent = = 9

Hence, b = 4 a

^{3}– 3 a + 5Now, y = 4 x

^{3}– 3x + 5dy/dx = 12 x

^{2}– 3Slope of the tangent =

Given that, slope of the tangent = slope of the perpendicular line

12a

^{2}– 3 = 912a

^{2}= 12a

^{2}= 1a = Â±1

So, b = 6 or 4.

Thus, slope of tangent = 9.

When a = 1 and b = 6, we have

y – y

_{1}= m (x – x_{1})y – 6 = 9 (x – 1)

y – 6 = 9x – 9

9x – y – 3 = 0

When a = -1 and b = 4, we have

y – y

_{1}= m (x – x_{1})y – 4 = 9 (x + 1)

y – 4 = 9x + 9

9x – y + 13 = 0

### Question 12. Find the equation of a normal to the curve y = x log_{e} x which is parallel to the line 2x âˆ’ 2y + 3 = 0.

**Solution:**

Slope of the given line is 1.

Let (a, b) be the point where the tangent is drawn to the curve.

Hence, b = a log

_{e}a . . . . (1)Now, y = x log

_{e}xdy/dx = x Ã— 1/x + log

_{e}x(1) = 1 + log_{e}x_{1}Slope of tangent = 1 + log a

Slope of normal =

Given that, slope of normal = slope of the given line.

=> -1 = 1 + log a

=> – 2 = log a

=> a = e

^{-2}From (1), we have

Now, b = e

^{-2}(-2) = -2 e^{-2}Given, (x

_{1}, y_{1}) = (e^{-2}, -2 e^{-2})The equation of normal is,

y + 2/e

^{2}= 1(x – 1/e^{2})y + 2/e

^{2}= x – 1/e^{2}x – y = 3/e

^{2}

### Question 13. Find the equation of the tangent line to the curve y = x^{2} âˆ’ 2x + 7

### (i) which is parallel to the line 2x âˆ’ y + 9 = 0?

**Solution:**

We have, y = x

^{2}âˆ’ 2x + 7On differentiating both sides, we get

dy/dx = 2x – 2

The equation of the line is 2x – y + 9 = 0

So the slope of line is 2.

According to the question,

=> 2x – 2 = 2

=> 2x = 4

=> x = 2

=> y = 2

^{2}âˆ’ 2(2) + 7 = 4 – 4 + 7 = 7As (x

_{1}, y_{1}) is (2, 7), the equation of tangent is,y – 7 = 2(x – 2)

y – 7 = 2x – 4

y – 2x – 3 = 0

### (ii) which is perpendicular to the line 5y âˆ’ 15x = 13.

**Solution:**

We have, y = x

^{2}âˆ’ 2x + 7On differentiating both sides, we get

dy/dx = 2x – 2

The equation of the line is 5y âˆ’ 15x = 13.

=> y = 3x + 13/5

So the slope of line is 3.

According to the question,

=> 2x – 2= -1/3

=> 6x – 6 = -1

=> x = 5/6

And y = 217/36.

As (x

_{1}, y_{1}) is (5/6, 217/36), the equation of tangent is,y – y

_{1}= m (x – x_{1})y – 217/36 = (-1/3) (x – 5/6)

36y -217 = -12x + 10

36y + 12x – 227 = 0

### Question 14. Find the equations of all lines having slope 2 and that are tangent to the curve y = 1/x – 3, x â‰ 3.

**Solution:**

Let (a , b) be the point where the tangent is drawn to this curve.

Since, the point lies on the curve, hence b = 1/(a – 3)

Slope of tangent, m =

Slope of the tangent} = 2

(a – 3)

^{2}= – 2a – 3 = âˆš-2, which does not exist because 2 is negative.

So, there does not exist any such tangent.

### Question 15. Find the equations of all lines of slope zero and that are tangent to the curve .

**Solution:**

Slope of the given tangent is 0.

Let (a, b) be a point where the tangent is drawn to the curve.

Since, the point lies on the curve, hence b = . . . . (1)

Slope of tangent =

Given that, slope of tangent = slope of the given line,

=> -2 a + 2 = 0

=> 2a = 2

=> a = 1

From (1), we get

Now, b =

(a, b) = (1, 1/2)

The equation of tangent is,

y – y

_{1}= m (x – x_{1})y – 1/2 = 0 (x – 1)

y = 1/2

### Question 16. Find the equation of the tangent to the curve which is parallel to the 4x âˆ’ 2y + 5 = 0.

**Solution:**

We have,

Let (a, b) be the point where the tangent is drawn to the curve y =

On differentiating both sides, we get

Slope of tangent at (a, b) =

Slope of line 4x âˆ’ 2y + 5 = 0 is 2.

Given that, slope of tangent = slope of the given line

9 = 16 (3a – 2)

9/16 = 3a – 2

3a = 9/16 + 2

a = 41/48

Now, b =

Therefore, (a, b) = (41/48, 3/4)

The equation of tangent is,

y – y

_{1}= m (x – x_{1})y – 3/4 = 2 (x – 41/48)

(4y – 3)/4 = 2 (48x – 41)/48

24y – 18 = 48x – 41

48x – 24y – 23 = 0

### Question 17. Find the equation of the tangent to the curve x^{2} + 3y âˆ’ 3 = 0, which is parallel to the line y= 4x âˆ’ 5.

**Solution:**

Suppose (a, b) be the required point.

We can find the slope of the given line by differentiating the equation w.r.t x,

So, slope of the line = 4

Since (a, b) lies on the curve, we get a

^{2}+ 3b âˆ’ 3 = 0 . . . . (1)Now,

2x + 3dy/dx = 0

dy/dx = -2x/3

Slope of tangent, m=

Given that tangent is parallel to the line, So we get,

Slope of tangent, m = slope of the given line

=> -2a/3 = 4

=> a = -6

From (1), we get

=> 36 + 3b – 3 = 0

=> 3b = – 33

=> b = – 11

(a, b) = (-6, -11)

The equation of tangent is,

y – y

_{1}= m (x – x_{1})y + 11 = 4 (x + 6)

y + 11 = 4x + 24

4x – y + 13 = 0

### Question 18. Prove that touches the straight line for all n âˆˆ N, at the point (a, b) ?

**Solution:**

We have,

On differentiating both sides, we get

Slope of tangent =

The equation of tangent is,

y – b = -b/a (x – a)

ya – ab = – xb + ab

xb + ya = 2ab

So, the given line touches the given curve at the given point.

Hence proved.

### Question 19. Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at t = Ï€/4.

**Solution:**

We have,

x = sin 3t, y = cos 2t

dx/dt = 3 cos3t and dy/dt = -2sin2t

Slope of tangent, m=

x

_{1}= sin 3Ï€/4 = 1/âˆš2 and y_{1}= cos Ï€/2 = 0So, (x

_{1}, y_{1}) = (1/âˆš2, 0)The equation of tangent is,

y – y

_{1}= m (x – x_{1})3y = 2âˆš2 x – 2

2âˆš2 x – 3y – 2 = 0

### Question 20. At what points will be tangents to the curve y = 2x^{3} âˆ’ 15x^{2} + 36x âˆ’ 21 be parallel to x-axis? Also, find the equations of the tangents to the curve at these points?

**Solution:**

We have,

y = 2x

^{3}âˆ’ 15x^{2}+ 36x âˆ’ 21The slope of x – axis is 0.

Let (a, b) be the required point.

Since (a, b) lies on the curve, we get

b = 2a

^{3}âˆ’ 15a^{2}+ 36a âˆ’ 21 . . . . (1)Also, we have

dy/dx = 6 x

^{2}– 30x + 36Slope of tangent at (a, b) =

Given that the slope of the tangent = slope of the x-axis, we have

=> 6a

^{2}– 30a + 36 = 0=> a

^{2}– 5a + 6 = 0=> (a – 2) (a – 3) = 0

=> a = 2 or a= 3

=> b = 7 or 6

When a = 2 and b = 7, the equation is,

y – y

_{1}= m (x – x_{1})y – 7 = 0 (x – 2)

y = 7

When a = 3 and b = 6, the equation is,

y – y

_{1}= m (x – x_{1})y – 6 = 0 (x – 3)

y = 6

### Question 21. Find the equation of the tangents to the curve 3x^{2} â€“ y^{2} = 8, which passes through the point (4/3, 0).

**Solution:**

We have,

3x

^{2}â€“ y^{2}= 8 . . . . (1)On differentiating both sides w.r.t x, we get

6x – 2y dy/dx = 0

2y dy/dx = 6x

dy/dx = 6x/2y

dy/dx = 3x/y

Let tangent at (h, k) pass through (4/3, 0). Since, (h, k) lies on (1), we get

3 h

^{2}– k^{2}= 8 . . . (ii)Slope of tangent at (h, k) = 3h/k

The equation of tangent at (h, k) is given by,

y – k = 3h/k (x – h)

Also,

=> 0 – k = (3h/k) (4/3 – h)

=> -k = 4h/k – 3h

^{2}/k=> – k

^{2}= 4h – 3h^{2}=> 8 – 3 h

^{2}= 4h – 3 h^{2}=> 8 = 4h

=> h = 2

Also we get,

=> 12 – k

^{2}= 8=> k

^{2}= 4=> k = Â±2

So, the points on curve (i) at which tangents pass through (4/3, 0) are (2, Â±2).

When h = 2 and k = 2, the equation is,

y – 2 = (6/2) (x – 2)

3x – y – 4 = 0

When h = 2 and k = â€“2, the equation is,

y + 2 = (6/-2) (x – 2)

3x + y – 4 = 0

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