RD Sharma Class 12 Ex 16.1 Solutions Chapter 16 Tangents and Normals

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter16
Exercise16.1
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 16.1 Solutions Chapter 16 Tangents and Normals

Question 1. Find the slopes of the tangent and the normal to the following curves at the indicated points:

(i) y = √x3 at x = 4

(ii) y = √x at x = 9

(iii) y = x3 – x at x = 2

(iv) y = 2x2 + 3 sin x at x = 0

(v) x = a(θ – sin θ), y = a(1 + cos θ) at θ = –π/2

(vi) x = a cos3 θ, y = a sin3 θ at θ = π/4

(vii) x = a(θ – sin θ), y = a(1 – cos θ) at θ = π/2

(viii) y = (sin 2x + cot x + 2)2 at x = π/2

(ix) x2 + 3y + y2 = 5 at (1, 1)

(x) xy = 6 at (1, 6)

Solution:

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

And slope of the normal = –1/slope of tangent = –1/(dy/dx).

(i) y = √x3 at x = 4

Differentiating y = √x3 with respect to x, we get,

Slope of tangent = 3x1/2/2

At x = 4, slope of tangent becomes,

Slope of tangent = 3(4)1/2/2 = 3(2)/2 = 3

And slope of normal at x = 4 is –1/3.

(ii) y = √x at x = 9

Differentiating y = √x with respect to x, we get,

Slope of tangent = x–1/2/2

At x = 9, slope of tangent becomes,

Slope of tangent = 9–1/2/2 = 1/[(3)(2)] = 1/6

And slope of normal at x = 9 is –6.

(iii) y = x3 – x at x = 2

Differentiating y = x3 – x with respect to x, we get,

Slope of tangent = 3x2 – 1

At x = 2, slope of tangent becomes,

Slope of tangent = 3(2)2 – 1 = 3(4) – 1 = 11

And slope of normal at x = 2 is –1/11.

(iv) y = 2x2 + 3 sin x at x = 0

Differentiating y = 2x2 + 3 sin x with respect to x, we get,

Slope of tangent = 4x + 3 cos x

At x = 0, slope of tangent becomes,

Slope of tangent = 4(0) + 3 cos 0 = 3.

And slope of normal at x = 0 is –1/3.

(v) x = a (θ – sin θ), y = a (1 + cos θ) at θ = –π/2

Differentiating x = a (θ – sin θ) with respect to θ, we get,

=> dx/dθ = a (1 – cos θ) . . . . (1)

Differentiating y = a (1 + cos θ) with respect to θ, we get,

=> dy/dθ = a (–sin θ) . . . . (2)

Dividing (2) by (1), we get,

dy/dx = Slope of tangent = –sin θ/(1 – cos θ)

At θ = –π/2, slope of tangent becomes,

Slope of tangent = –sin (–π/2)/(1 – cos (–π/2))

= 1/(1–0)

= 1

And slope of normal at θ = –π/2 is –1.

(vi) x = a cos3 θ, y = a sin3 θ at θ = π/4

Differentiating x = a cos3 θ with respect to θ, we get,

=> dx/dθ = a [(3cos2 θ) (–sin θ)]

= –3a cos2 θ sin θ . . . . (1)

Differentiating y = a sin3 θ with respect to θ, we get,

=> dy/dθ = a [(3sin2 θ) (cos θ)]

= 3a sin2 θ cos θ . . . . (2)

Dividing (2) by (1), we get,

dy/dx = Slope of tangent =\frac{3asin^2θcosθ}{-3acos^2θsinθ} = – tan θ

At θ = π/4, slope of tangent becomes,

Slope of tangent = – tan (π/4)

= −1

And slope of normal at θ = π/4 is 1.

(vii) x = a (θ – sin θ), y = a (1 – cos θ) at θ = π/2

Differentiating x = a (θ – sin θ) with respect to θ, we get,

=> dx/dθ = a (1 – cos θ) . . . . (1)

Differentiating y = a (1 – cos θ) with respect to θ, we get,

=> dy/dθ = a (sin θ) . . . . (2)

Dividing (2) by (1), we get,

dy/dx = Slope of tangent = sin θ/(1−cosθ)

= – tan θ

At θ = π/2, slope of tangent becomes,

Slope of tangent = sin π/2/(1−cos π/2)

= 1/(1−0)

= 1

And slope of normal at θ = π/2 is −1.

(viii) y = (sin 2x + cot x + 2)2 at x = π/2

Differentiating y = (sin 2x + cot x + 2)2 with respect to x, we get,

Slope of tangent = 2 (sin 2x + cot x + 2) (2 cos 2x – cosec2 x)

At x = π/2, slope of tangent becomes,

Slope of tangent = 2 (sin 2(π/2) + cot π/2 + 2) (2 cos 2(π/2) – cosec2 (π/2))

= 2 (0 + 0 + 2) (–2 – 1)

= –12

And slope of normal at x = π/2 is 1/12.

(ix) x2 + 3y + y2 = 5 at (1, 1)

Differentiating x2 + 3y + y2 = 5 with respect to x, we get,

=> 2x + 3 (dy/dx) + 2y (dy/dx) = 0

=> 2x + dy/dx (2y+3) = 0

=> Slope of tangent = dy/dx = –2x/(2y+3)

At x = 1 and y = 1, slope of tangent becomes,

Slope of tangent = –2(1)/[2(1)+3] = –2/5

And slope of normal at (1, 1) is 5/2.

(x) xy = 6 at (1, 6)

Differentiating xy = 6 with respect to x, we get,

=> x (dy/dx) + y = 0

=> Slope of tangent = dy/dx = –y/x

At x = 1 and y = 6, slope of tangent becomes,

Slope of tangent = –6/1 = –6

And slope of normal at (1, 6) is 1/6.

Question 2. Find the values of a and b if the slope of the tangent to the curve xy + ax + by = 2 at (1, 1) is 2.

Solution:

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

Differentiating xy + ax + by = 2 with respect to x, we get

=> x (dy/dx) + y + a + b (dy/dx) = 2

=> dy/dx = −(a+y)/(x+b)

As we are given dy/dx = 2, we get,

=> −(a+y)/(x+b) = 2

Now at x = 1 and y = 1, we get,

=> −(a+1)/(1+b) = 2

=> −a − 1 = 2 + 2b

=> a + 2b = –3 . . . . (1)

Now the point (1, 1) also lies on the curve, so we have,

=> 1 × 1 + a × 1 + b × 1 = 2

=> 1 + a + b = 2

=> a + b = 1 . . . . (2)

Subtracting (1) from (2), we get,

=> –b = 1+3

=> b = –4

Putting b = –4 in (1), we get,

=> a = 1+4 = 5

Therefore, the value of a is 5 and b is –4.

Question 3. If the tangent to the curve y = x3 + ax + b at (1, –6) is parallel to the line x – y + 5 = 0, find a and b.

Solution:

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

Differentiating y = x3 + ax + b with respect to x, we get

=> dy/dx = 3x2 + a

Now at x = 1 and y = –6, we get,

=> dy/dx = 3(1)2 + a

=> dy/dx = 3 + a . . . . (1)

Now this curve is parallel to the line x – y + 5 = 0.

=> y = x + 5

Therefore slope of the line is 1. So, the slope of the curve will also be 1 as slope of parallel lines are equal. So, from (1), we get,

=> dy/dx = 1 . . . . (2)

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

=> a + 3 = 1

=> a = –2 . . . . (3)

Now at x = 1 and y = –6, our curve y = x3 + ax + b becomes,

=> –6 = 1 + a + b

=> a + b = –7

Using (3), we get,

=> b = –7 – (–2)

=> b = –5

Therefore, the value of a is –2 and b is –5.

Question 4. Find a point on the curve y = x3 – 3x where the tangent is parallel to the chord joining (1, – 2) and (2, 2).

Solution:

We are given the coordinates of the chord (1, – 2) and (2, 2).

Therefore, slope of the chord =\frac{2-(-2)}{2-1} = 4

Given curve is y = x3 – 3x. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 3x2 – 3

As the tangent is parallel to the chord, its slope must be equal to 4.

=> 3x2 – 3 = 4

=> 3x2 = 7

=> x =\pm\sqrt{\frac{7}{3}}

Putting value of x in the curve y = x3 – 3x, we get

=> y = x (x2 – 3)

=> y =\pm\sqrt{\frac{7}{3}}(\frac{7}{3}-3)

=> y =\pm\frac{-2}{3}\sqrt{\frac{7}{3}}

Therefore, the required point is(\pm\sqrt{\frac{7}{3}},\pm\frac{-2}{3}\sqrt{\frac{7}{3}}) .

Question 5. Find a point on the curve y = x3 – 2x2 – 2x at which the tangent lines are parallel to the line y = 2x – 3.

Solution:

Given curve is y = x3 – 2x– 2x. We know that the slope of the tangent of a curve is given by dy/dx.

=> dy/dx = 3x2 – 4x – 2 . . . . (1)

Now this curve is parallel to the line y = 2x – 3 whose slope is 2. So, the slope of the curve will also be 2. So, from (1), we get,

=> 3x2 – 4x – 2 = 2

=> 3x2 – 6x + 2x – 4 = 0

=> 3x (x – 2) + 2 (x – 2) = 0

=> (x – 2) (3x + 2) = 0

=> x = 2 or x = –2/3

If x = 2, we get

y = (2)3 – 2 × (2)2 – 2 × (2)

= 8 – 8 – 4

= – 4

And if x = –2/3, we get,

y = (–2/3)3 – 2 × (–2/3)2 – 2 × (–2/3)

=\frac{-8-24+36}{27}

= 4/27

Therefore, (2, –4) and (–2/3, 4/27) are the required points.

Question 6. Find a point on the curve y2 = 2x3 at which the slope of the tangent is 3.

Solution:

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

=> 2y dy/dx = 6x2

=> dy/dx = 3x2/y . . . . (1)

As it is given that the slope of tangent is 3, we get,

=> 3x2/y = 3

=> 3x2 = 3y

=> x2 = y

Putting this in the curve y2 = 2x3, we get,

=> (x2)2 = 2x3

=> x4 − 2x3 = 0

=> x3 (x − 2) = 0

=> x = 0 or x = 2

If x = 0, we get, y = 0. Putting these values in (1), we get dy/dx = 0, which is not possible as the given value of slope is 3.

And if x =2, we get y = 4.

Therefore, the required point is (2, 4).

Question 7. Find a point on the curve xy + 4 = 0 at which the tangents are inclined at an angle of 45o with the x–axis.

Solution:

Given curve is xy + 4 = 0. We know that the slope of the tangent of a curve is given by dy/dx.

=> x (dy/dx) + y = 0

=> dy/dx = −y/x . . . . (1)

We are given that the tangent is inclined at an angle of 45o with the x–axis. So slope of the tangent is,

dy/dx = tan 45o = 1.

So, (1) becomes,

=> −y/x = 1

=> y = −x

Putting this in the curve xy + 4 = 0 we get,

=> x(−x) + 4 = 0

=> x2 = 4

=> x = ±2

When x = 2, y = −2.

And when x = −2, y = 2.

Therefore, the required points are (2, – 2) & (– 2, 2).

Question 8. Find a point on the curve y = xwhere the slope of the tangent is equal to the x – coordinate of the point.

Solution:

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

=> dy/dx = 2x . . . . (1)

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

Therefore, dy/dx = x . . . . (2)

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

2x = x

=> x = 0

Putting this in the curve y = x2, we get,

=> y = 02

=> y = 0

Therefore, the required point is (0, 0).

Question 9. At what point on the circle x2 + y2 – 2x – 4y + 1 = 0, the tangent is parallel to x – axis.

Solution:

Given circle is x2 + y2 – 2x – 4y + 1 = 0. We know that the slope of the tangent of a curve is given by dy/dx.

=> 2x + 2y (dy/dx) – 2 – 4 (dy/dx) = 0

=> dy/dx = (1– x)/(y– 2)

As the tangent is parallel to x-axis, its slope is equal to 0.

So, (1– x)/(y– 2) = 0

=> x = 1

Putting x = 1 in the circle x2 + y2 – 2x – 4y + 1 = 0, we get,

=> 1 + y2 – 2 – 4y +1 = 0

=> y2 – 4y = 0

=> y (y – 4) = 0

=> y = 0 and y = 4

Therefore, the required points are (1, 0) and (1, 4).

Question 10. At what point of the curve y = x2 does the tangent make an angle of 45o with the x–axis?

Solution:

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

=> dy/dx = 2x . . . . (1)

We are given that the tangent is inclined at an angle of 45o with the x–axis. So slope of the tangent is

Therefore, dy/dx = tan 45o = 1 . . . . (2)

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

2x = 1

=> x = 1/2

Putting this in the curve y = x2, we get,

=> y = (1/2)2

=> y = 1/4

Therefore, the required point is (1/2, 1/4).

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 = x− 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 x+ y− 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
=> x− 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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