# Class 12 RD Sharma Solutions – Chapter 16 Tangents and Normals – Exercise 16.1 | Set 2

**Question 11. Find the points on the curve y = 3x**^{2} âˆ’ 9x + 8 at which the tangents are equally inclined with the axes.

^{2}âˆ’ 9x + 8 at which the tangents are equally inclined with the axes.

**Solution:**

Given curve is y = 3x

^{2}âˆ’ 9x + 8. We know that the slope of the tangent of a curve is given by dy/dx.=> dy/dx = 6x âˆ’ 9 . . . . (1)

We are given that the tangent is equally inclined with the axes. So Î¸ = Ï€/4 or â€“Ï€/4.

Hence, slope of the tangent is Â±1.

=> 6x âˆ’ 9 = 1 or 6x âˆ’ 9 = â€“1

=> 6x = 10 or 6x = 8

=> x = 5/3 or x = 4/3

When x = 5/3,

y = 3 (5/3)

^{2}âˆ’ 9 (5/3) + 8 = 4/3When x = 4/3,

y = 3 (4/3)

^{2}âˆ’ 9 (5/3) + 8 = 4/3

Therefore, the required points are (5/3, 4/3) and (4/3, 4/3).

**Question 12. At what points on the curve y = 2x**^{2} âˆ’ x + 1 is the tangent parallel to the line y = 3x + 4?

^{2}âˆ’ x + 1 is the tangent parallel to the line y = 3x + 4?

**Solution:**

Given curve is y = 2x

^{2}âˆ’ x + 1. We know that the slope of the tangent of a curve is given by dy/dx.=> dy/dx = 4x âˆ’ 1 . . . . (1)

We are given that the tangent is parallel to the line y = 3x + 4. Now the slope of the line is 3, so slope of tangent must also be 3. So, we have,

=> 4x âˆ’ 1 = 3

=> x = 1

Putting x = 1 in the curve y = 2x

^{2}âˆ’ x + 1, we gety = 2(1) âˆ’ 1 + 1 = 2

Therefore, the required point is (1, 2).

**Question 13. Find the point on the curve y = 3x**^{2} + 4 at which the tangent is perpendicular to the line whose slope is âˆ’1/6.

^{2}+ 4 at which the tangent is perpendicular to the line whose slope is âˆ’1/6.

**Solution:**

Given curve is y = 3x

^{2}+ 4. We know that the slope of the tangent of a curve is given by dy/dx.=> dy/dx = 6x

It is given that the tangent is perpendicular to the line whose slope is âˆ’1/6. So the product of both the slopes must be âˆ’1.

Therefore the slope of tangent, dy/dx = 6.

=> 6x = 6

=> x = 1

Putting x = 1 in the curve y = 3x

^{2}+ 4, we get,=> y = 3(1)

^{2}+ 4 = 3 + 4 = 7

Therefore, (1, 7) is the required point.

**Question 14. Find the points on the curve x**^{2} + y^{2} = 13, the tangent at each one of which is parallel to the line 2x + 3y = 7.

^{2}+ y

^{2}= 13, the tangent at each one of which is parallel to the line 2x + 3y = 7.

**Solution:**

Given curve is x

^{2}+ y^{2}= 13. We know that the slope of the tangent of a curve is given by dy/dx.=> 2x + 2y dy/dx = 0

=> dy/dx = âˆ’x/y . . . . (1)

It is given that the tangent is parallel to the line 2x + 3y = 7.

=> 3y = âˆ’2x + 7

=> y = âˆ’(2/3)x + 7/3

Therefore slope of the line is âˆ’2/3 and slope of the tangent is also âˆ’2/3 as slope of parallel lines are equal.

=> dy/dx = âˆ’2/3 . . . . (2)

From (1) and (2), we get,

=> âˆ’x/y = âˆ’2/3

=> x = 2y/3 . . . . (3)

Putting x = 2y/3 in the curve x

^{2}+ y^{2}= 13, we get,=> 4y

^{2}/9 + y^{2}= 13=> 13y

^{2}/9 = 13=> y

^{2}= 9=> y = Â±3

Putting y = Â±3, in (3), we get,

When y = 3, x = 2 and when y = âˆ’3, x = âˆ’2.

Therefore, the required points are (2, 3) and (âˆ’2, âˆ’3).

**Question 15. Find the points on the curve 2a**^{2}y = x^{3} âˆ’ 3ax^{2} where the tangent is parallel to x-axis.

^{2}y = x

^{3}âˆ’ 3ax

^{2}where the tangent is parallel to x-axis.

**Solution:**

Given curve is 2a

^{2}y = x^{3}âˆ’ 3ax^{2}. We know that the slope of the tangent of a curve is given by dy/dx.=> 2a

^{2}dy/dx = 3x^{2}âˆ’ 3a (2x)=> dy/dx =

It is given that the tangent is parallel to x-axis, so the slope of the tangent becomes 0.

=> = 0

=> 3x (x âˆ’ 2a) = 0

=> x = 0 or x = 2a

When x = 0, the value of y from the curve is,

=> y =

=> y =

=> y = 0

And when x = 2a, the value of y is,

=> y =

=> y =

=> y = âˆ’2a

Therefore, the required points are (0, 0) and (2a, âˆ’2a).

**Question 16. At what points on the curve y = x**^{2 }âˆ’ 4x + 5 is the tangent perpendicular to the line 2y + x = 7?

^{2 }âˆ’ 4x + 5 is the tangent perpendicular to the line 2y + x = 7?

**Solution:**

Given curve is y = x

^{2}âˆ’ 4x + 5. We know that the slope of the tangent of a curve is given by dy/dx.=> dy/dx = 2x âˆ’ 4 . . . . (1)

It is given that the tangent is perpendicular to the line 2y + x = 7.

=> 2y = âˆ’x + 7

=> y = âˆ’(1/2)x + 7/2

Therefore slope of the line is âˆ’1/2 and product of this slope with that of tangent is âˆ’1 as both lines are perpendicular to each other.

So, slope of tangent is 2.

=> dy/dx = 2 . . . . (2)

From (1) and (2), we get,

=> 2x âˆ’ 4 = 2

=> x = 3

Putting this in the curve y = x

^{2}âˆ’ 4x + 5, we get=> y = x2 âˆ’ 4x + 5

= (3)

^{2}âˆ’ 4(3) + 5= 2

Therefore, the required point is (3, 2).

**Question 17. Find points on the curve x**^{2}/4 + y^{2}/25 = 1 at which the tangents are

^{2}/4 + y

^{2}/25 = 1 at which the tangents are

**(i) parallel to the x-axis**

**Solution:**

Given curve is x

^{2}/4 + y^{2}/25 = 1. We know that the slope of the tangent of a curve is given by dy/dx.=> 2x/4 + 2y/25 (dy/dx) = 0

=> dy/dx = âˆ’25x/4y

As it is given that the tangent is parallel to the x-axis, its slope must be 0.

=> âˆ’25x/4y = 0

=> x = 0

Putting this in the curve x

^{2}/4 + y^{2}/25 = 1, we get=> y

^{2}= 25=> y = Â±5

Therefore, the required points are (0, 5) and 0, âˆ’5).

**(ii) parallel to the y-axis**

**Solution:**

Slope of the tangent = dy/dx = âˆ’25x/4y

Therefore, slope of the normal = = 4y/25x

As it is given that the tangent is parallel to the y-axis, the slope of the normal must be 0.

=> 4y/25x = 0

=> y = 0

Putting this in the curve x

^{2}/4 + y^{2}/25 = 1, we get=> x

^{2}= 4=> x = Â±2

Therefore, the required points are (2, 0) and (âˆ’2, 0).

**Question 18. Find the points on the curve x**^{2} + y^{2} âˆ’ 2x âˆ’ 3 = 0 at which the tangents are parallel to

^{2}+ y

^{2}âˆ’ 2x âˆ’ 3 = 0 at which the tangents are parallel to

**(i) x-axis**

**Solution:**

Given curve is x

^{2 }+ y^{2 }âˆ’ 2x âˆ’ 3 = 0. We know that the slope of the tangent of a curve is given by dy/dx.=> 2x + 2y (dy/dx) âˆ’ 2 = 0

=> dy/dx = (1âˆ’x)/y

As it is given that the tangent is parallel to the x-axis, its slope must be 0.

=> (1âˆ’x)/y = 0

=> x = 1

Putting this in the curve x

^{2}+ y^{2}âˆ’ 2x âˆ’ 3 = 0, we get=> 1 + y

^{2}âˆ’ 2 âˆ’ 3 = 0=> y

^{2}= 4=> y = Â±2

Therefore, the required points are (1, 2) and (1, âˆ’2).

**(ii) y-axis **

**Solution:**

Slope of the tangent = dy/dx = (1âˆ’x)/y

Therefore, slope of the normal = = y/(xâˆ’1)

As it is given that the tangent is parallel to the y-axis, the slope of the normal must be 0.

=> y/(xâˆ’1) = 0

=> y = 0

Putting this in the curve x

^{2}+ y^{2}âˆ’ 2x âˆ’ 3 = 0, we get=> x

^{2 }âˆ’ 2x âˆ’ 3 = 0=> x = âˆ’1, 3

Therefore, the required points are (âˆ’1, 0) and (3, 0).

**Question 19. Find points on the curve x**^{2}/9 + y^{2}/16 = 1 **at which the tangents are**

^{2}/9 + y

^{2}/16 = 1

**(i) parallel to the x-axis **

**Solution:**

Given curve is x

^{2}/9 + y^{2}/16 = 1. We know that the slope of the tangent of a curve is given by dy/dx.=> 2x/9 + 2y/16 (dy/dx) = 0

=> dy/dx = âˆ’16x/9y

As it is given that the tangent is parallel to the x-axis, its slope must be 0.

=> âˆ’16x/9y = 0

=> x = 0

Putting this in the curve x

^{2}/9 + y^{2}/16 = 1, we get=> y

^{2}= 16=> y = Â±4

Therefore, the required points are (0, 4) and 0, âˆ’4).

**(ii) parallel to the y-axis**

**Solution:**

Slope of the tangent = dy/dx = âˆ’16x/9y

Therefore, slope of the normal = = 9y/16x

As it is given that the tangent is parallel to the y-axis, the slope of the normal must be 0.

=> 9y/16x = 0

=> y = 0

Putting this in the curve x

^{2}/9 + y^{2}/16 = 1, we get=> x

^{2}= 9=> x = Â±3

Therefore, the required points are (3, 0) and (âˆ’3, 0).

**Question 20. Show that the tangents to the curve y = 7x**^{3} + 11 at the points where x = 2 and x = âˆ’2 are parallel.

^{3}+ 11 at the points where x = 2 and x = âˆ’2 are parallel.

**Solution:**

Given curve is y = 7x

^{3}+ 11. We know that the slope of the tangent of a curve is given by dy/dx.=> dy/dx = 21x

^{2}Now slope at x = 2 is

=> dy/dx = 21(2)

^{2}= 84And slope at x = âˆ’2 is,

=> dy/dx = 21(âˆ’2)

^{2}= 84As the slopes at x = 2 and x = âˆ’2 are equal, these tangents are parallel.

Hence proved.

**Question 21. Find the points on the curve y = x**^{3} where the slope of the tangent is equal to the x-coordinate of the point.

^{3}where the slope of the tangent is equal to the x-coordinate of the point.

**Solution:**

Given curve is y = x

^{3}. We know that the slope of the tangent of a curve is given by dy/dx.=> dy/dx = 3x

^{2}It is given that the slope of the tangent is equal to the x-coordinate of the point.

=> 3x

^{2}= x=> x(3x âˆ’ 1) = 0

=> x = 0 or x = 1/3

When x = 0, y = 0

^{3}= 0And when x = 1/3, y = (1/3)

^{3}= 1/27

Therefore, the required points are (0, 0) and (1/3, 1/27).