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

• Last Updated : 07 Jul, 2022

### Question 11. Find the points on the curve y = 3x2 âˆ’ 9x + 8 at which the tangents are equally inclined with the axes.

Solution:

Given curve is y = 3x2 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 = 4/3

When 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 = 2x2 âˆ’ x + 1 is the tangent parallel to the line y = 3x + 4?

Solution:

Given curve is y = 2x2 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 = 2x2 âˆ’ x + 1, we get

y = 2(1) âˆ’ 1 + 1 = 2

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

### Question 13. Find the point on the curve y = 3x2 + 4 at which the tangent is perpendicular to the line whose slope is âˆ’1/6.

Solution:

Given curve is y = 3x2 + 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 = 3x2 + 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 x2 + y2 = 13, the tangent at each one of which is parallel to the line 2x + 3y = 7.

Solution:

Given curve is x2 + y2 = 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 x2 + y2 = 13, we get,

=> 4y2/9 + y2 = 13

=> 13y2/9 = 13

=> y2 = 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 2a2y = x3 âˆ’ 3ax2 where the tangent is parallel to x-axis.

Solution:

Given curve is 2a2y = x3 âˆ’ 3ax2. We know that the slope of the tangent of a curve is given by dy/dx.

=> 2a2 dy/dx = 3x2 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 = x2 âˆ’ 4x + 5 is the tangent perpendicular to the line 2y + x = 7?

Solution:

Given curve is y = x2 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 = x2 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 x2/4 + y2/25 = 1 at which the tangents are

(i) parallel to the x-axis

Solution:

Given curve is x2/4 + y2/25 = 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 x2/4 + y2/25 = 1, we get

=> y2= 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 x2/4 + y2/25 = 1, we get

=> x2= 4

=> x = Â±2

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

### Question 18. Find the points on the curve x2 + y2 âˆ’ 2x âˆ’ 3 = 0 at which the tangents are parallel to

(i) x-axis

Solution:

Given curve is x2 + y2 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 x2 + y2 âˆ’ 2x âˆ’ 3 = 0, we get

=> 1 + y2 âˆ’ 2 âˆ’ 3 = 0

=> y2 = 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 x2 + y2 âˆ’ 2x âˆ’ 3 = 0, we get

=> x2 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 x2/9 + y2/16 = 1at which the tangents are

(i) parallel to the x-axis

Solution:

Given curve is x2/9 + y2/16 = 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 x2/9 + y2/16 = 1, we get

=> y2= 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 x2/9 + y2/16 = 1, we get

=> x2= 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 = 7x3 + 11 at the points where x = 2 and x = âˆ’2 are parallel.

Solution:

Given curve is y = 7x3 + 11. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 21x2

Now slope at x = 2 is

=> dy/dx = 21(2)2 = 84

And slope at x = âˆ’2 is,

=> dy/dx = 21(âˆ’2)2 = 84

As 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 = x3 where the slope of the tangent is equal to the x-coordinate of the point.

Solution:

Given curve is y = x3. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 3x2

It is given that  the slope of the tangent is equal to the x-coordinate of the point.

=> 3x2 = x

=> x(3x âˆ’ 1) = 0

=> x = 0 or x = 1/3

When x = 0, y = 03 = 0

And when x = 1/3, y = (1/3)3 = 1/27

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

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