Chapter 16 of RD Sharma's Class 12 Mathematics textbook focuses on the tangents and normals to the curves a key concept in calculus. Exercise 16.2 | Set 2 delves into solving problems related to finding the equations of the tangents and normals to given curves at specific points. This exercise enhances understanding of how to apply derivative concepts to the geometrical problems involving curves.
Chapter 16 Tangents and Normals - Exercise 16.1 | Set 2
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 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).
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Summary
Exercise 16.1 (Set 2) in Chapter 16 (Tangents and Normals) of RD Sharma's Class 12 mathematics textbook focuses on finding the equations of tangents and normals to curves at specific points. The problems in this set require students to apply the concept of differentiation to determine the slope of the tangent line and then use it to derive the equation of the tangent and normal. This exercise helps students develop a strong understanding of the relationship between the derivative of a function and the properties of its tangent and normal lines.
What is the main objective of Exercise 16.1 (Set 2)?
The main objective of this exercise is to help students find the equations of tangents and normals to curves at specific points using the concepts of differentiation.
What is the relationship between the derivative of a function and the slope of the tangent line?
The slope of the tangent line to a curve at a point is equal to the derivative of the function at that point.
How do you find the equation of the tangent line given the point of tangency and the slope of the tangent line?
To find the equation of the tangent line, you can use the point-slope form of the equation: y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope of the tangent line.
How do you find the equation of the normal line given the point of tangency and the slope of the tangent line?
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Once you have the slope of the normal line, you can use the point-slope form to find its equation.
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