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

  • Last Updated : 30 Apr, 2021

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

Solution:

Given curve is y = 3x2 − 9x + 8. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 6x − 9  . . . . (1)

We are given that the tangent is equally inclined with the axes. So θ = π/4 or –π/4. 

Hence, slope of the tangent is ±1. 



=> 6x − 9 = 1 or 6x − 9 = –1

=> 6x = 10 or 6x = 8

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

When x = 5/3,

y = 3 (5/3)2 − 9 (5/3) + 8 = 4/3

When x = 4/3,

y = 3 (4/3)2 − 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 − x + 1 is the tangent parallel to the line y = 3x + 4?

Solution:

Given curve is y = 2x2 − x + 1. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 4x − 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 − 1 = 3

=> x = 1

Putting x = 1 in the curve y = 2x2 − x + 1, we get

y = 2(1) − 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 −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 −1/6. So the product of both the slopes must be −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 = −x/y  . . . . (1)

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

=> 3y = −2x + 7

=> y = −(2/3)x + 7/3

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

=> dy/dx = −2/3  . . . . (2)

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

=> −x/y = −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 = ±3

Putting y = ±3, in (3), we get,

When y = 3, x = 2 and when y = −3, x = −2. 

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

Question 15. Find the points on the curve 2a2y = x3 − 3ax2 where the tangent is parallel to x-axis.

Solution:



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

=> 2a2 dy/dx = 3x2 − 3a (2x)

=> dy/dx = \frac{3x^2-6ax}{2a^2}

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

=> \frac{3x^2-6ax}{2a^2}   = 0

=> 3x (x − 2a) = 0

=> x = 0 or x = 2a

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

=> y = \frac{x^3-3ax^2}{2a^2}

=> y = \frac{0-0}{2a^2}



=> y = 0

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

=> y = \frac{(2a)^3-3a(2a)^2}{2a^2}

=> y = \frac{8a^3-12a^3}{2a^2}

=> y = −2a

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

Question 16. At what points on the curve y = x2 − 4x + 5 is the tangent perpendicular to the line 2y + x = 7? 

Solution:

Given curve is y = x2 − 4x + 5. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 2x − 4   . . . . (1)

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

=> 2y = −x + 7

=> y = −(1/2)x + 7/2

Therefore slope of the line is −1/2 and product of this slope with that of tangent is −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 − 4 = 2

=> x = 3

Putting this in the curve y = x2 − 4x + 5, we get

=> y = x2 − 4x + 5 

= (3)2 − 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 = −25x/4y

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

=> −25x/4y = 0



=> x = 0

Putting this in the curve x2/4 + y2/25 = 1, we get

=> y2= 25

=> y = ±5

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

(ii) parallel to the y-axis

Solution:

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

Therefore, slope of the normal = \frac{-1}{\frac{-25x}{4y}}   = 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 = ±2

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

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

(i) x-axis

Solution:

Given curve is x2 + y2 − 2x − 3 = 0. We know that the slope of the tangent of a curve is given by dy/dx.

=> 2x + 2y (dy/dx) − 2 = 0

=> dy/dx = (1−x)/y

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

=> (1−x)/y = 0

=> x = 1

Putting this in the curve x2 + y2 − 2x − 3 = 0, we get

=> 1 + y2 − 2 − 3 = 0

=> y2 = 4

=> y = ±2 

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

(ii) y-axis 

Solution:

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

Therefore, slope of the normal = \frac{-1}{\frac{(1−x)}{y}}   = y/(x−1)

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

=> y/(x−1) = 0

=> y = 0

Putting this in the curve x2 + y2 − 2x − 3 = 0, we get

=> x2 − 2x − 3 = 0

=> x = −1, 3

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



Question 19. Find points on the curve x2/9 + y2/16 = 1 at 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 = −16x/9y

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

=> −16x/9y = 0

=> x = 0

Putting this in the curve x2/9 + y2/16 = 1, we get

=> y2= 16

=> y = ±4

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

(ii) parallel to the y-axis

Solution:

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

Therefore, slope of the normal = \frac{-1}{\frac{-16x}{9y}}   = 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 = ±3

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

Question 20. Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = −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 = −2 is,

=> dy/dx = 21(−2)2 = 84

As the slopes at x = 2 and x = −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 the slope of the tangent is equal to the x-coordinate of the point. 

=> 3x2 = x

=> x(3x − 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).

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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