The availability of step-by-step ML Aggarwal Maths for Class 12 Solutions Chapter 7 Applications of Derivatives Ex 7.2 can make challenging problems more manageable.
ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.2
Question 1.
What is the slope of the tangent to the following curves :
(i) y = x3 – x at x = 2? (NCERT)
(ii) y = 3x4 – 4x at x = 4? (NCERT)
(iii) x2 + 3y + y2 = 5 at the point (1, 1)?
(iv) y = x3 – 3x + 2 at the point whose x coordinate is 3? (NCERT)
(y) y = 3x2 + 4x at the point whose x coordinate is – 2 ?
(vi) y = 2 sin2 (3x) at \(\frac{1}{3}\)?
Solution:
(i) Given equation of curve be y = x3 – x
Diff. both sides w.r.t. x, we get
\(\frac{d y}{d x}\) = 3x2 – 1
∴ (\(\frac{d y}{d x}\))x = 2 = 3 × 22 – 1
= 12 – 1 = 11
Thus, slope of tangent at x = 2.
= (\(\frac{d y}{d x}\))x = 2 = 11
(ii) Given curve is y = 3x4 – 4x
∴ \(\frac{d y}{d x}\) = 12x3 – 4
∴ (\(\frac{d y}{d x}\))x = 4 = 12 (4)3 – 4 = 764.
(iii) Given eqn. of curve be x3 + 3y + y2 – 5 = 0
Diff. both sides w.r.t. x ; we get
2x + 3 \(\frac{d y}{d x}\) + 2y \(\frac{d y}{d x}\) = 0
⇒ (3 + 2y) \(\frac{d y}{d x}\) = – 2x
⇒ \(\frac{d y}{d x}\) = \(\frac{-2 x}{2 y+3}\)
\(\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{-2}{2+3}=\frac{-2}{5}\)
∴ slope of tangent at (1, 1) = (\(\frac{d y}{d x}\))(1, 1)
= \(-\frac{2}{5}\)
(iv) Given curve be, y = x3 – 3x + 2
∴ \(\frac{d y}{d x}\) = 3x2 – 3
∴ (\(\frac{d y}{d x}\))x = 3 = 3 (3)2 – 3 = 24
(v) Given, eqn. of curve be y = 3x2 + 4x
Diff. both sides w.r.t. x, we have
∴ Slope of tangent to given curve at x = – 2 = (\(\frac{d y}{d x}\))x = – 2
= 6 × (- 2) + 4
= – 12 + 4 = – 8.
(vi) Given y = sin2 (3x)
∴ \(\frac{d y}{d x}\) = 4 sin 3x . cos 3x . 3
= 12 sin 3x cos 3x
∴ slope of tangent to given curve at (x = \(\frac{\pi}{6}\)) = (\(\frac{d y}{d x}\))x = \(\frac{\pi}{6}\)
= 12 × sin \(\frac{\pi}{2}\) cos \(\frac{\pi}{2}\) = 0.
Question 2.
What is the slope of the normal to the following curves :
(i) y = x3 – 5x2 – x + 1 at the point(1, – 4)?
(ii) y = 2x2 + 3 sin x at x = 0? (NCERT)
Solution:
(i) Given eqn. of curve be,
y = x3 – 5x2 – x + 1 …………….(1)
∴ \(\frac{d y}{d x}\) = 4x2 – 10x – 1
Thus, the slope of the normal to curve (1) at point (1, – 4) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(1,-4)}}\)
= \(=\frac{-1}{3-10-1}\)
= \(\frac{-1}{-8}=\frac{1}{8}\)
(ii) Given eqn. of curve be,
y = 2x2 + 3 sin x
∴ \(\frac{d y}{d x}\) = 4x + 3 cos x
Thus (\(\frac{d y}{d x}\))x = 0 = 4 × 0 + 3 × 1 = 3
∴ slope of normal to curve (1) at x = 0 = – \(\frac{1}{\left(\frac{d y}{d x}\right)_{x=0}}\)
= – \(\frac{1}{3}\).
Question 3.
Find the point on the curve y = 3x2 – 2x + 1 at which the slope of the tangent is 4.
Solution:
Let P (x1, y1) be any point on given curve.
Eqn. of given curve be, y = 3x2 – 2x + 1 ………..(1)
∴ \(\frac{d y}{d x}\) = 6x – 2
∴ slope of tangent to the curve (1) at P(x1, y1)
= 6x1 – 2
also, given slope of tangent to given curve = 4
∴ 6x1 – 2 = 4
⇒ x1 = 1
Now P (x1, y1) also lies on given curve (1)
y1 = 3x12 – 2x1 + 1
y1 = 3(1)2 – 2x1 + 1
= 4 – 2 = 2
Thus, the required point on curve be (1, 2).
Question 4.
Determine the point on the curve y = 3x2 – 1 at which the slope of the tangent is 3.
Solution:
Let P (x1, y1) be any point on given curve
and eqn. of given curve be. y = 3x2 – 1
∴ \(\frac{d y}{d x}\) = 6x
Thus, slope of tangent to given curve at (x1, y1) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 6x1
also slope of tangent to given curve = 3
∴ 6x1 = 3
⇒ x1 = \(\frac{1}{2}\)
Since, the point (x1 , y1) be lies on given curve
∴ y1 = 3x12 – 1
⇒ y1 = 3 × \(\frac{1}{4}\) – 1
= \(\frac{1}{4}\)
Thus, the required point on the given curve be \(\left(\frac{1}{2},-\frac{1}{4}\right)\).
Question 5.
At what point on the curve y = x2 does the tangent make an angle of 45° with the x-axis?
Solution:
Let the eqn. of given curve be y = x2 ………..(1)
Let the required point on the given curve be (x1, y1)
∴ y1 = x12 …………()
∴ \(\frac{d y}{d x}\) = 2x
⇒ (\(\frac{d y}{d x}\))(x1, y1) = 2x1
It is given that,
slope of tangent at (x1, y1) = tan \(\frac{\pi}{4}\)
∴ (\(\frac{d y}{d x}\))(x1, y1) = 1
⇒ 2x1 = 1
⇒ x1 = \(\frac{1}{2}\)
∴ from (2) ;
y1 = \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
Hence the required point be \(\left(\frac{1}{2}, \frac{1}{4}\right)\).
Question 6.
Find the point on the curvey = x2 – 2x + 3 at which the tangent is parallel to x-axis.
Solution:
Eqn. of given curve y = x2 – 2x + 3 ……….(1)
Let P (x, y) be any point on curve (1)
∴ \(\frac{d y}{d x}\) = 2x – 2
Now slope of tangent to given curve (1) is || to x-axis
\(\frac{d y}{d x}\) = 0
⇒ 2x – 2 = 0
⇒ x = 1
∴ from (1) ;
y = 12 – 2 × 1 + 3 = 2
Thus required point on given curve be (1, 2).
Question 6 (old).
Find the slope of the tangent to the curve y = \(\frac{x-1}{x-2}\) at x = 10. (NCERT)
Solution:
Given eqn. of curve be, y = \(\frac{x-1}{x-2}\)
∴ \(\frac{d y}{d x}=\frac{x-2-(x-1)}{(x-2)^2}\)
= \(\frac{-1}{(x-2)^2}\) ; x ≠ 2.
∴ \(\left.\frac{d y}{d x}\right]_{x=10}=\frac{-1}{64}\)
Hence, slope of tangent to given curve at x = 10.
= \(\left(\frac{d y}{d x}\right)_{x=10}=\frac{-1}{64}\).
Question 7.
Find the slope of the normal to the curve x = a cos3 θ, y = a sin3 θ at θ = \(\frac{\pi}{4}\). (NCERT)
Solution:
Given eqn. of curve be x = a cos3 θ ………..(1)
and y = a sin3 θ ……..(2)
Difi. eqn. (1) and (2) w.r.t. θ ; we have
\(\frac{d x}{d \theta}\) = 3a cos2 θ (- sin θ)
\(\frac{d y}{d \theta}\) = 3a sin2 θ cos θ
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\)
= \(\frac{3 a \sin ^2 \theta \cos \theta}{-3 a \cos ^2 \theta \sin \theta}\)
= – tan θ
Thus slope of normal to given curve at θ = \(\frac{\pi}{4}\)
= \(\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{4}}\)
= \( \frac{-1}{-\tan \frac{\pi}{4}}\)
= 1.
Question 8.
Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos2 θ at θ = \(\frac{\pi}{2}\).
Solution:
Given eqn. of curve be
x = 1 – a sin θ …………(1)
and y = b cos2 θ …………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. θ ; we have
\(\frac{d x}{d \theta}\) = – a cos θ ………..(3)
\(\frac{d y}{d \theta}\) = 2b cos θ (- sin θ)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}\)
= \(\frac{-2 b \cos \theta \sin \theta}{-a \cos \theta}\)
= \(\frac{2 b}{a}\) sin θ
Thus slope of normal to given curve at θ = \(\frac{\pi}{2}\) is given by = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{2}}}\)
= \(\frac{-1}{\frac{2 b}{a} \sin \frac{\pi}{2}}\)
= \(-\frac{a}{2 b}\)
Question 9.
Write the equation of the normal to the curve y2 = 8x at the origin.
Solution:
given eqn. of curve be, y2 = 8x
∴ 2y \(\frac{d y}{d x}\) = 8
⇒ \(\frac{d y}{d x}\) = \(\frac{4}{y}\)
∴ slope of normal to curve at (0, 0) = – \(\frac{1}{\left(\frac{d y}{d x}\right)_{(0,0)}}\)
= \(-\frac{1}{\infty}\) = 0
Thus, eqn. of normal to given curve at (0, 0) be given by
y – 0 = – \(\frac{1}{\left(\frac{d y}{d x}\right)}\) (x – 0)
⇒ y – 0 = 0
⇒ y = 0.
Question 10.
Find the slopes of the tangents at the three points where the curve y = x (x2 – 3x – 4) cuts the x-axis.
Solution:
Given eqn. of curve be
y = x (x2 – 3x – 4) …………….(1)
eqn. (1) meets x-axis i.e. y = 0
Thus, 0 = x (x2 – 3x – 4)
⇒ 0 = x (x + 1) (x – 4)
⇒ x = 0, – 1, 4
Hence, the curve meets x-axis at (0, 0) ; (- 1, 0) and (4, 0).
Diff. eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = 3x2 – 6x – 4
∴ Slope of tangent to given curve at (0, 0) = \(\left(\frac{d y}{d x}\right)_{(0,0)}\)
= 0 – 0 – 4 = – 4
slope of tangent to given curve at (- 1, 0) = \(\left(\frac{d y}{d x}\right)_{(-1,0)}\)
= 3 (- 1)2 – 6 (- 1) – 4 = 5
and slope of tangent to given curve at (4, 0) = \(\left(\frac{d y}{d x}\right)_{(4,0)}\)
= 3 (4)2 – 6 × 4 – 4
= 48 – 24 – 4 = 20.
Question 11.
Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = – 2 are yy parallel. (NCERT)
Solution:
Given curve is y = 7x3 + 11
∴ \(\frac{d y}{d x}\) = 21 x2
∴ slope of tangent at x = 2 = \(\left.\frac{d y}{d x}\right]_{x=2}\) = 84
slope of tangent at x = 2 = \(\left.\frac{d y}{d x}\right]_{x=-2}\) = 84
Since both slopes are equal.
∴ tangents to the curve at x = 2 and – 2 are parallel.
Question 12.
If the tangent to tile curve y = x3 + ax + b at (1, – 6) is parallel to the line 2x – 2y + 7 = 0, find a and b.
Solution:
Given eqn. of given curve be y = x3 + ax + b ………(1)
Diff. both sides of eqn. (1) w.r.t. x;
\(\frac{d y}{d x}\) = 3x2 + a
∴ \(\left(\frac{d y}{d x}\right)_{(1,-6)}\) = 3 + a
it is given that slope of tangent to curve (1) at (1, – 6) slope of line x – y + 5 = 0
∴ \(\left(\frac{d y}{d x}\right)_{(1,-6)}=\frac{-1}{-1}\)
⇒ 3 + a = 1
⇒ a = – 2
Also the point (1, – 6) lies on curve (1),
∴ – 6 = 13 + a + b
⇒ a + b = – 7
⇒ – 2 + b = – 7
⇒ b = – 5.
Question 13.
Determine the point of the curve y = 3x2 – 5 at which the tangent is perpendicular to a line whose slope is – \(\frac{1}{3}\).
Solution:
Let P (x1, y1) be any point on given curve
y = 3x2 – 5 ………..(1)
∴ y1 = 3x12 – 5 ………..(2)
Diff. eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = 6x
∴ slope of tangent to given curve at (x1, y1) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= m1 = 6x1
Also tangent ⊥to line whose slope is = – \(\frac{1}{3}\)
∴ m2 = – \(\frac{1}{3}\)
Thus, m1m2 = – 1
⇒ 6x1 (- \(\frac{1}{3}\)) = – 1
⇒ – 2x1 = – 1
⇒ x1 = \(\frac{1}{2}\)
∴ from (2) ; we have
y1 = 3 (\(\frac{1}{2}\))2 – 5
= \(\frac{3}{4}\) – 5
= – \(\frac{17}{4}\)
Thus, the required point on given curve be \(\left(\frac{1}{2},-\frac{17}{4}\right)\).
Question 14.
Determine the points on the curve y = x3 – 3x2 – 9x + 7 at which the tangents are Parallel to x-axis. (NCERT)
Solution:
Given eqn. of curve be
y = x3 – 3x2 – 9x + 7 ………….(1)
Let P (x1, y1) be any point on given curve (1).
∴ y = x13 – 3x12 – 9x1 + 7 ………..(2)
Diff. (1) both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = 3x2 – 6x – 9
Thus, the slope of the tangent to given curve at (x1, y1) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 3x12 – 6x1 – 9
Since the tangent are parallel to x-axis.
∴ slope of tangent is 0.
Thus \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\) = 0.
⇒ 3 (x12 – 2x1 – 3) = 0
⇒ (x1 + 1) (x1 – 3) = 0
⇒ x1 = – 1, 3
When x1 = – 1
∴ from (2) ; we have
y1 = – 1 – 3 + 9 + 7 = 12
when x1 = 3
∴ from (2) ; we have
y1 = 27 – 27 – 27 + 7 = – 20
Thus, the required points on given curve be (- 1, 12) and (3, – 20).
Question 15.
Find the points on the curve x2 + y2 – 2x – 3 = 0 at which the tangents are parallel to x-axis.
Solution:
Let the eqn. of given curve be
x2 + y2 – 2x – 3 = 0 ……….(1)
Let the required point on given curve (1) be (x1, y1).
∴ x12 + y12 – 2x1 – 3 = 0 ………..(2)
Diff. eqn. (1) w.r.t. x ; we have
2x + 2y \(\frac{d y}{d x}\) – 2 = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{1-x}{y}\)
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{1-x_1}{y_1}\)
Since it is given that tangent is parallel to x-axis.
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\) = 0
⇒ \(\frac{1-x_1}{y_1}\) = 0
⇒ x1 = 1
∴ from (2) ;
1 + y12 – 2 – 3 = 0
⇒ y1 = ± 2
Hence the required points are (1, ± 2).
Question 16.
Find the points on the curve \(\frac{x^2}{4}+\frac{y^2}{25}\) = 1 at which the tangents are
(a) parallel to x-axis
(b) parallel to y-axis. (NCERT)
Solution:
(a) Given curve is,
\(\frac{x^2}{4}+\frac{y^2}{25}\) = 1 …………..(1)
Differentiate both sides w.r.t. x, we get
∴ \(\frac{2 x}{4}+\frac{2 y}{25} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{25 x}{4 y}\)
Since tangent is parallel to x-axis.
∴ \(\frac{d y}{d x}\) = 0
⇒ \(\frac{- 25 x}{4 y}\) = 0
⇒ x = 0
∴ From (1) ;
y = ± 5
Here required points are (0, ± 5).
(b) Since tangent is || to y-axis
∴ \(\frac{d y}{d x}\) = ∞
⇒ \(\frac{d y}{d x}\) = 0
∴ y = 0
∴ From (1) ; x = ± 2.
Hence required points are (± 2, 0).
Question 17.
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:
Eqn. of given curve be x2 + y2 = 13 ………….(1)
Let the required point be on given curve (1) be (x1, y1).
∴ x12 + y12 = 13 ………..(2)
Since the tangent is parallel to given line 2x + 3y = 7
∴ slope of tangent at (x1, y1) = slope of line 2x + 3y = 7
⇒ \(\left(\frac{d y}{d x}\right)_{x_1 y_1}=-\frac{2}{3}\) …………(3)
Diff. eqn. (1) w.r.t. x; we get
2x + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}-\frac{x}{y}\)
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1 y_1\right)}=-\frac{x_1}{y_1}\)
∴ From (3) ; we have
⇒ \(-\frac{x_1}{y_1}=-\frac{2}{3}\)
⇒ 3x1 = 2y1
∴ from (2) ;
x12 + \(\frac{9}{4}\) x12 = 13
⇒ \(\frac{13}{4}\) x12 = 13
⇒ x1 = ± 2
When x1 = 2
∴ from (4) ;
y1 = 3
When x1 = – 2
∴ from (4) ;
y1 = – 3
Hence the required points are (2, 3) and (- 2, – 3).
Question 18.
(Find the points on the curve y = x3 – 3x2 + 2x at which the tangent lines are parallel to the line y – 2x + 3 = 0.
Solution:
Given eqn. of curve be y = x3 – 3x2 + 2x …………(1)
Let P (x1, y1) be any point on given curve (1).
∴ y1 = x13 – 3x12 + 2x1 ………….(2)
Diff. both sides of eqn. (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = 3x2 – 6x + 2
∴ slope of tangent to given curve (1) at (x1, y1) = 3x12 – 6x1 + 2
slope of given Line y – 2x + 3 = 0
= – \(\frac{(-2)}{1}\) = 2
Since the tangent to given curve is parallel to given line
∴ 3x12 – 6x1 + 2 = 2
⇒ 3x1 (x1 – 2) = 0
⇒ x1 = 0, 2
When x1 = 0
∴ from (2) ; we have
y1 = 0 – 0 + 0 = 0
When x1 = 2
∴ from (2); we have
y1 = 8 – 12 + 4 = 0
Thus the required points on given curve be (0, 0) and (2, 0).
Question 19.
Find the equations of the tangent and the normal to each of the following curves at the given point :
(i) y = x3 at (1,1)
(ii) y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3) (NCERT)
(iii) y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5) (NCERT)
(iv) y = x2 at (0, 0) (NCERT)
(v) y = sin2 x at x = \(\frac{\pi}{2}\)
(vi) \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 at (x0, y0). (NCERT)
Solution:
(i) Given eqn. of curve be y = x3 ………..(1)
∴ \(\frac{d y}{d x}\) = 3x2
∴ slope of tangent to curve (1) at given point (1, 1) = \(\left(\frac{d y}{d x}\right)_{(1,1)}\)
= 3 × 12 = 3
Thus, eqn. of tangent to given curve at (1, 1) is given by
y – 1 = 3 (x – 1)
⇒ 3x – y – 2 = 0
Slope of normal to given curve at (1, 1) = \(-\frac{1}{\left(\frac{d y}{d x}\right)_{(1,1)}}=-\frac{1}{3}\)
∴ required eqn. of normal to given curve (1) at(1, 1) is given by
y – 1 = – \(\frac{1}{3}\) (x – 1)
⇒ x + 3y —4 = 0.
(ii) Given,
y = x4 – 6x3 + 13x2 – 10x + 5
∴ \(\frac{d y}{d x}\) = 4x3 – 18x2 + 26x – 10
when x = 1
∴ from (1) we have
y = 1 – 6 + 13 – 10 + 5 = 3
∴ slope of tangent at (1, 3) = \(\left(\frac{d y}{d x}\right)_{(1,3)}\)
= 4— 18 + 26 – 10 = 2
and slope of Normal at (1, 3) = – \(\frac{1}{2}\)
∴ eqn. of tangent at (1, 3) is given by
y – 3 = 2 (x – 1)
⇒ 2x – y + 1 = 0
and eqn. of Normal at (1, 3) is given by y – 3 = – \(\frac{1}{2}\) (x – 1)
⇒ x + 2y – 7 = 0.
(iii) Given curve be y = x4 – 6x3 + 13x2 – 10x + 5
∴ \(\frac{d y}{d x}\) = 4x3 – 18x2 + 26x – 10
Thus slope of tangent at (0, 5) = \(\left(\frac{d y}{d x}\right)_{(0,5)}\) = – 10
∴ slope of normal at (0, 5) = \(-\frac{1}{\left(\frac{d y}{d x}\right)_{(0,5)}}\)
= \(-\frac{1}{-10}=\frac{1}{10}\)
Thus eqn. of tangent at (0, 5) is given by y – 5 = – 10 (x – 0)
⇒ y + 10x – 5 = 0
and eqn. of Normal to curve at (0, 5) is given by y – 5 = \(\frac{1}{10}\) (x – 0)
⇒ 10y – 50 = x
⇒ x – 10y + 50 = 0.
(iv) Given curve be, y = x2
∴ \(\frac{d y}{d x}\) = 2x
∴ slope of tangent at (0, 0) = \(\left(\frac{d y}{d x}\right)_{(0,0)}\)
= 2 × 0 = 0
and slope of normal at (0, 0) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(0,0)}}\) = ∞
∴ eqn. of tangent at (0, 0) is y – 0 = \(\frac{-1}{\left(\frac{d y}{d x}\right)}\) (x – 0) = 0
and eqn. of normal at (0, 0) is y – 0 = \(\frac{-1}{\left(\frac{d y}{d x}\right)}\) (x – 0)
⇒ x = 0.
(v) Given curve be y = sin2x
∴ \(\frac{d y}{d x}\) = 2 sin x cos x = sin 2x
∴ slope of tangent at x = \(\frac{\pi}{2}\)
= \(\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}}\) = 0
and slope of normal at x = \(\frac{\pi}{2}\) = ∞
When x = \(\frac{\pi}{2}\)
⇒ y = sin2 \(\frac{\pi}{2}\)
= (1)2 = 1
∴ eqn. of tangent to given curve at given point (\(\frac{\pi}{2}\), 1) is
y – 1 = 0 (x – \(\frac{\pi}{2}\))
⇒ y – 1 = 0
and eqn. of normal at (\(\frac{\pi}{2}\), 1) is
y – 1 = \(\frac{-1}{\left(\frac{d y}{d x}\right)}\) (x – \(\frac{\pi}{2}\))
⇒ x – \(\frac{\pi}{2}\) = 0.
Question 19 (old).
Find the equations of all lines having slope 2 and being tangents to the curve y + \(\frac{2}{x-3}\) = 0. (NCERT)
Solution:
Given eqn. of curve be y + \(\frac{2}{x-3}\) = 0
⇒ y = – \(\frac{2}{x-3}\) …………….(1)
Let the point (x1, y1) be on curve at which slope of tangent be equal to 2.
∴ \(\frac{d y}{d x}=\frac{2}{(x-3)^2}\)
Thus slope of tangent to given curve (1) at (x1, y1) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= \(\frac{2}{\left(x_1-3\right)^2}\)
\(\frac{2}{\left(x_1-3\right)^2}\) = 2
⇒ (x1 – 3)2 = 1
⇒ x1 – 3 = ± 1
⇒ x1 = 3 ± 1
⇒ x1 = 4, 2
Also the point (x1, y1) lies on eqn. (1)
∴ y1 = – \(\frac{2}{x_1-3}\) …………..(2)
When x1 = 4
∴ from (2) ; y1 = \(\frac{-2}{4-3}\) = – 2
When x1 = 2
∴ from (2) ; y1 = \(\frac{-2}{2-3}\) = 2
Hence the points on given curve be (4, – 2) and (2, 2).
∴ eqn. of line having slope 2 and passes through (4, – 2) is given by
y + 2 = 2 (x – 4)
⇒ 2x – y – 10 = 0
and the eqn. of line through (2, 2) and having slope 2 is given by
y – 2 = 2 (x – 2)
⇒ 2x – y – 2 = 0
Question 20.
Find the equation of the normal to the curve ay2 = x3 at the point (am2, am3)
Solution:
The given curve be, ay2 = x3
∴ 2ay \(\frac{d y}{d x}\) = 3x2
⇒ \(\frac{d y}{d x}\) = \(\frac{3 x^2}{2 a y}\)
⇒ \(\left.\frac{d y}{d x}\right]_{\left(a m^2, a m^3\right)}=\frac{3\left(a m^2\right)^2}{2 a\left(a m^3\right)}\)
= \(\frac{3 a^2 m^4}{2 a^2 m^3}=\frac{3}{2} m\)
∴ slope of tangent at (am2, am3) = \(\frac{3 m}{2}\)
∴ slope of normal at (am2, am3) = \(\frac{-1}{\frac{3 m}{2}}=\frac{-2}{3 m}\)
∴ eqn. of Normal at (am2, am3) is given by y – am3 = \(\frac{-2}{3 m}\) (x – am2)
⇒ 3my – 3am4 = – 2x + 2am2
⇒ 2x + 3my = 3am4 + 2am2.
Question 21.
(i) Find the equation of the tangent to the curve √x + √y = a at the point \(\left(\frac{a^2}{4}, \frac{a^2}{4}\right)\).
(ii) Find the equation of the tamgent and the normal to the curve x2/3 + y2/3 = 2 at (1, 1) (NCERT)
Solution:
(i) Equation of the curve is, √x + √y = a …………(1)
Differentiate eqn. (1) w.r.t. x, we have
\(\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{\sqrt{y}}{\sqrt{x}}\)
slope of tangent at \(\left.\left(\frac{a^2}{4}, \frac{a^2}{4}\right)=\frac{d y}{d x}\right]\left(\frac{a^2}{4}, \frac{a^2}{4}\right)\)
= \(-\frac{a / 2}{a / 2}\) = – 1
Thus, eqn. of tangent at \(\left(\frac{a^2}{4}, \frac{a^2}{4}\right)\) is given by
y – \(\frac{a^2}{4}\) = – 1 (x – \(\frac{a^2}{4}\))
⇒ x + y – \(\frac{a^2}{2}\) = 0.
(ii) Given, eqn. of the curve x2/3 + y2/3 = 2 ……….(1)
Diff. both sides of eqn. (1) w.r.t. x ; we have
\(\frac{2}{3} x^{-\frac{1}{3}}+\frac{2}{3} y^{-\frac{1}{3}} \frac{d y}{d x}\) = 0
⇒ \(\frac{1}{y^{1 / 3}} \frac{d y}{d x}=-\frac{1}{x^{1 / 3}}\)
⇒ \(\frac{d y}{d x}=-\frac{y^{1 / 3}}{x^{1 / 3}}\)
\(\left(\frac{d y}{d x}\right)_{(1,1)}=-\frac{1}{1}\) = – 1
The required eqn. of tangent at (1, 1) is given by
y – 1 = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) (x – 1)
⇒ y – 1 = – 1 (x – 1)
⇒ y – 1 = – x + 1
⇒ x + y = 2
The required eqn. of Normal at (1, 1) is given by y – 1 = – \(\frac{1}{\left(\frac{d y}{d x}\right)_{(1,1)}}\) (x – 1)
⇒ y – 1 = 1 (x – 1)
⇒ y – x = 0.
Question 21 (old).
Find the equations of the tangents to the curve y = x3 + 2x – 4 which are perpendicular to the line x + 14y +3 = 0. (NCERT)
Solution:
Given eqn. of curve be y = x3 + 2x – 4 ………..(1)
Let (x1, y1) be any point on the curve (1) at which tangents are to be drawn.
∴\(\frac{d y}{d x}\) = 3x2 + 2
Thus slope of tangent to curve (1) at (x1, y1) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 3x12 + 2
Also it is given that. tangent is ⊥ to the given line. x + 14y + 3 = 0, whose slope is – \(\frac{1}{14}\).
Thus, (3x12 + 2) (- \(\frac{1}{14}\)) = – 1
⇒ 312 + 2 = 14
⇒ 312 = 12
⇒ x12 = 4
⇒ x1 = ± 2
Since the point (x1, y1) lies on given curve (1).
∴ y1 = x1 + 2x1 – 4 ……….(2)
When x1= 2
∴ from (2) ; we have
y1 = 8 + 4 – 4 = 8
When x1 = – 2
∴ from (2) ; we have
y1 = – 8 – 4 – 4 = – 16
∴ required points on given curve are (2, 8) and (- 2, – 16).
Thus eqn. of tangent at (2, 8) is given by y – 8 = 14 (x – 2)
⇒ 14x – y – 20 = 0
and eqn. of tangent at (- 2,- 16) is given by
y + 16 = 14 (x + 2)
⇒ 14x – y + 12 = 0.
Question 22.
(i) Find the equation of tangent to the curve given by x = a sin3 t, y = b cos3 t at a point where t = \(\frac{\pi}{2}\). (NCERT)
(ii) Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at t = \(\frac{\pi}{4}\).
Solution:
(i) Given eqn. of curve be,
x = a sin3 t …………..(1)
and y = b cos3 t …………..(2)
Diff. eqn. (1) and eqn. (2) w.r.t. t, we have
\(\frac{d x}{d t}\) = 3a sin2 t cos t
\(\frac{d y}{d t}\) = 3b cos2 t (- sin t)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{3 b \cos ^2 t(-\sin t)}{3 a \sin ^2 t \cos t}\)
= – \(\frac{b}{a}\) cot t
Thus slope of tangent to given curve at t = \(\frac{\pi}{2}\)
= \(\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{2}}\)
= – \(\frac{b}{a}\) cot \(\frac{\pi}{2}\) = 0
Therefore, the required eqn. of tangent at x = a sin3 \(\frac{\pi}{2}\) = a
and y = b cos3 \(\frac{\pi}{2}\) = 0
i.e. (a, 0) is given by y – 0 = 0 (x – a)
⇒ y = 0.
(ii) Given curve be x = sin 3t,
y = cos 2t ………..(1)
∴ \(\frac{d x}{d t}\) = 3 cos 3t
and \(\frac{d y}{d t}\) = – 2 sin 2t
Thus, \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)
= – \(\frac{2 \sin 2 t}{3 \cos 3 t}\)
∴ slope of tangent at t = \(\left.\frac{\pi}{4} \frac{d y}{d x}\right]\)
= \(-\frac{2}{3} \frac{\sin \frac{\pi}{2}}{\cos \frac{3 \pi}{4}}\)
= \(-\frac{2}{3} \cdot \frac{1}{-\frac{1}{\sqrt{2}}}\)
= \(\frac{2 \sqrt{2}}{3}\)
Also from (1) ;
when t = \(\frac{\pi}{4}\)
⇒ x = \(\frac{1}{\sqrt{2}}\) and y = 0
eqn. of tangent at \(\left(\frac{1}{\sqrt{2}}, 0\right)\), is given by y – 0 = \(\frac{2 \sqrt{2}}{3}\left(x-\frac{1}{\sqrt{2}}\right)\)
⇒ 2√2x – 3y – 2 = 0.
Question 23.
(i) Find the equations of all lines having slope 2 and that are tangents to the curve y = \(\frac{1}{x-3}\). (NCERT)
(ii) Find the equation of the tangent to the curve y = x + \(\frac{4}{x^2}\) which is parallel to x-axis.
Solution:
(i) Given curve is y = \(\frac{1}{x-3}\), x ≠ 3
∴ \(\frac{d y}{d x}=\frac{2}{(x-3)^2}\)
According to given condition, we have \(\frac{2}{(x-3)^2}\) = 2 which is not possible
L.H.S. is -ve and R.H.S. is +ve.
∴ There is no such tangent is possible whose slope is 2.
(ii) Given equation of curve be
y = x + \(\frac{4}{x^2}\) ………….(1)
∴ Slope of tangent to given curve (1) at any point (x, y) = \(\frac{d y}{d x}\)
= 1 – \(\frac{8}{x^3}\)
Since tangent is parallel to x-axis
∴ \(\frac{d y}{d x}\) = 0
⇒ 1 – \(\frac{8}{x^3}\) = 0
⇒ x = 2
⇒ from (1) ;
y = 2 + \(\frac{4}{2^2}\) = 3
Hence any point on curve (1) be (2, 3).
Thus, the required equation of tangent to given curve (1) at point (2, 3) and parallel to x-axis is given by
y – 3 = 0 (x – 2)
⇒ y = 3.
Question 24.
Find the equation of the tangent to the curve y = x2 – 2x + 7 which is
(i) parallel to the line 2x – y + 9 = 0
(ii) perpendicular to the line 5y – 15x = 13. (NCERT)
Solution:
(i) eqn. of given curve be
y = x2 – 2x + 7 ……………(1)
∴ \(\frac{d y}{d x}\) = 2x – 2
Thus slope of tangent = \(\frac{d y}{d x}\) = 2x – 2
But tangent is || to the line 2x – y + 9 = 0, whose slope is 2.
∴ 2x – 2 = 2 [∵ m1 = m2]
∴ from (1) ;
y = 4 – 4 + 7 = 7
Hence point of contact is (2, 7).
∴ eqn. of tangent at (2, 7) is given by
y – 7 = 2(x – 2)
⇒ 2x – y + 3 = 0.
(ii) But the tangent is ⊥ to line 5y – 15x – 13 = 0 whose slope is 3.
∴ product of slopes = – 1
⇒ (2x – 2) (3) = – 1
⇒ 6x = 5
⇒ x = \(\frac{5}{6}\)
∴ from (1) ;
y = \(\frac{25}{36}-\frac{10}{6}\) + 7
⇒ y = \(\frac{25-60+252}{36}=\frac{217}{36}\)
∴ required point of contact is \(\left(\frac{5}{6}, \frac{217}{36}\right)\).
Thus eqn. of tangent at \(\left(\frac{5}{6}, \frac{217}{36}\right)\) is given by
y – \(\frac{217}{26}\) = – \(\frac{1}{3}\left(x-\frac{5}{6}\right)\)
⇒ 36 y – 217 = – 12x + 10
⇒ 12x + 36y = 227.
Question 25.
(i) Find the equations of tangents to the curve y = 4x3 – 3x + 5 which are perpendicular to the line 9y + x + 3 =0.
(ii) Find the equations of the tangents to the curve 3x2 – y2 = 2 which are perpendicular to the line x + 3y = 2.
Solution:
(i) Given, eqn. of curve be, y = 4x3 – 3x + 5 …………(1)
∴ \(\frac{d y}{d x}\) = 12x2 – 3
∴ slope of tangent to given curve = \(\left(\frac{d y}{d x}\right)_{(x, y)}\)
= 12x2 – 3
∴ eqn. of given line be 9y + x + 3 = 0 ………..(2)
∴ slope of eqn. (1) = – \(\frac{1}{9}\)
Since tangent to curve (2) is ⊥ to line (2).
∴ (12x2 – 3) (- \(\frac{1}{9}\)) = – 1
⇒ 12x2 – 3 = 9
⇒ x2 = 1
⇒ x = ± 1
∴ from (1) ;
y = 4 – 3 + 5 = 6
and y = – 4 + 3 + 5 = 4
Thus point on given curves are (1, 6) and (- 1, 4)
Hence slope of tangent to given curve (1) = \(\left(\frac{d y}{d x}\right)_{x= \pm 1}\)
= 12 – 3 = 9.
Thus eqn. of tangent to curve (1) at point (1, 6) be given by
y – 6 = 9 (x – 1)
⇒ 9x – y – 3 = 0
and eqn. of tangent to curve (1) at point (- 1, 4) be given by
y – 4 = 9 (x + 1)
⇒ 9x – y +1 2 = 0
(ii) Given, eqn. of curve be 3x2 – y2 = 2 ……….(1)
Let P (x1, y1) be point on the curve, at which tangent to the curve are drawn.
Difff. eqn. (1) w.r.t. x, we have
6x – 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{6 x}{2 y}=\frac{3 x}{y}\)
Thus slope of tangent to curve (1) at (x1, y1) = \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= \(\frac{3 x_1}{y_1}\)
Also slope of given line x + 3y – 2 = 0 be – \(\frac{1}{3}\).
Since the tangent to given curve is ⊥ to given line
∴ \(\left(\frac{3 x_1}{y_1}\right)\left(-\frac{1}{3}\right)\) = – 1
[∵ m1 m2 = – 1]
⇒ x1 = y1 ………..(2)
Also the point (x1, y1) lies on given curve (1).
∴ 3x12 – y12 = 2
3x12 – x12 = 2 [using (2)]
⇒ 2x12 = 2
⇒ x12 = ± 1
∴ from(2) ;
y1 = ± 1
Thus, the points on given curve be (1, 1) and (- 1, – 1) and slope of tangent to curve be equal to 3.
Therefore, the eqn. of tangent to given curve at (1, 1) is given by
y – 1 = 3 (x – 1)
⇒ 3x – y – 2 = 0
and eqn. of tangent to curve at (- 1, – 1) is given by
y + 1 = 3 (x + 1)
⇒ 3x – y + 2 = 0.
Question 26.
Find the equations of the normals to the une 3x2 – y2 = 8 which are parallel to the line x + 3y = 4. (NCERT Exemplar)
Solution:
Given eqn. of curve be 3x2 – y2 = 8 ……….(1)
Diff. eqn. (1) w.r.t. x, we have
6x – 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=\frac{3 x}{y}\)
Hence slope of tangent = \(\frac{d y}{d x}=\frac{3 x}{y}\)
∴ slope of normal = – \(\frac{y}{3 x}\)
But normal is || to line x + 3y = 4, whose slope is – \(\frac{1}{3}\).
∴ \(\frac{-y}{3 x}=-\frac{1}{3}\) [∵ m1 = m2]
⇒ y = x
∴ from (1);
3x2 – x2 = 8
⇒ x2 = 4
⇒ x = ± 2
When x = 2, from (1) ;
y2 = 4
⇒ y = ± 2
Hence points on curve are (2, 2) and (2, – 2)
When x = – 2;
from(1);
y2 = 4
y = ± 2
and pts on curve are (- 2, 2) and (- 2, – 2)
Thus eqn. of normal at (2, 2) is
y – 2 = – (x – 2)
⇒ x + 3y – 8 = 0
eqn. of normal at (2, – 2) is
y + 2 = – (x – 2)
⇒ x + 3y + 4 = 0
eqn. of normal at (- 2, 2) is
y – 2 = – \(\frac{1}{3}\) (x + 2)
⇒ x + 3y – 4 = 0
and eqn. of normal at (- 2, – 2) is
y + 2 = – \(\frac{1}{3}\) (x + 2)
⇒ x+ 3y+ 8 = 0.
Question 27.
Find the equation of be normal to the curve y = x + \(\frac{1}{x}\), x > 0, perpendicular to the line 3x – 4y = 7.
Solution:
Given eqn. of curve be y = x + \(\frac{1}{x}\) ………….(1)
∴ \(\frac{d y}{d x}\) = 1 – \(\frac{1}{x^2}\)
eqn. of given line be,
3x – 4y – 7 = 0 …………(2)
∴ slope of line (2) = \(\frac{-3}{-4}=\frac{3}{4}\)
∴ slope of normal to given curve (1) = – \(\frac{1}{\left(\frac{d y}{d x}\right)_{(x, y)}}\)
Since normal to curve (1) is ⊥ to line (2)
– \(\left(\frac{1}{1-\frac{1}{x^2}}\right)\left(\frac{3}{4}\right)\) = – 1
⇒ 1 – \(\frac{1}{x^2}\) = \(\frac{3}{4}\)
⇒ \(\frac{1}{x^2}=\frac{1}{4}\)
⇒ x = ± 2
but x > 0
∴ x = 2
∴ from (1) ;
y = 2 + \(\frac{1}{2}\) = \(\frac{5}{2}\)
Thus any point on given curve (1) be (2, \(\frac{5}{2}\))
and slope of normal = – \(\frac{1}{\left(\frac{d y}{d x}\right)_{\left(2, \frac{5}{2}\right)}}\)
= \(\frac{-1}{1-\frac{1}{4}}=-\frac{4}{3}\)
Hence eqn. of normal to given curve at (2, \(\frac{5}{2}\)) be given by
y – \(\frac{5}{2}\) = – \(\frac{4}{3}\) (x – 2)
⇒ 6y – 15 = – 8x + 16
⇒ 8x + 6y – 31 = 0.
Question 28.
Find the equation of the normal to the curve x2 = 4y which passes through the point (- 1, 4).
Solution:
eqn. of given curve be x2 = 4y …………(1)
Let the normal at P (x1, y1) on curve (1) pass through the point (- 1, 4).
Diff. (1) w.r.t. x; we have
2x = 4 \(\frac{d y}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{x}{2}\)
∴ slope of tangent to curve at P = (\(\frac{d y}{d x}\))P
= \(\frac{x_1}{2}\)
Also point P lines on curve (1)
∴ x12 = 4y1 …………..(2)
Thus eqn. of normal to given curve (1) at P(x1, y1) be given by
y – y1 = – \(\frac{1}{\left(\frac{d y}{d x}\right)_P}\) (x – x1)
⇒ y – y1 = – \(\frac{2}{x_1}\) (x – x1) ………..(3)
eqn. (3) passes through point (- 1, 4)
∴ 4 – y1 = – \(\frac{2}{x_1}\) (- 1 – x1)
⇒ 4 – y1 = \(\frac{2}{x_1}\) + 2
⇒ y1 = 2 – \(\frac{2}{x_1}\)
∴ from (2) ;
x12 = 4 (2 – latex]\frac{2}{x_1}[/latex])
⇒ x13 = 8x1 – 8
⇒ x13– 8x1 + 8 = 0
⇒ x1 = 2
∴ from (2) ;
y1 = 1
Thus eqn. of normal to given curve at point (2, 1) be given by
y – 1 = – \(\frac{1}{1}\) (x – 2)
⇒ x + y – 3 = 0.
Question 29.
Find the angle of intersection between the following curves :
(i) y = 6 and x2y = 12
(ii) y=4 – x2and y = x2 (NCERT Exemplar)
(iii) y = x and x2 = y. (NCERT Exemplar)
Solution:
(i) The eqn’s of given curves are xy = 6 ………..(1)
and x2y = 12 ………..(2)
For point of intersection of eqn. (1) and eqn. (2),
we have to solve eqn. (1) and (2) simultaneously, we get
From (2) ;
x . (xy) = 12
⇒ 6x = 12 [using (1)]
⇒ x = 2
∴ from (1) ;
2y = 6
⇒ y = 3
Thus the point of intersection of two given curves be (2, 3).
Diff. eqn (1) and eqn. (2) w.r.t. x ; we have
x + \(\frac{d y}{d x}\) y = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{y}{x}\)
∴ slope of tangent to curve (1) at (2, 3) = m1
= \(\left(\frac{d y}{d x}\right)_{(2,3)}=\frac{-3}{2}\)
Also, x2 \(\frac{d y}{d x}\) + 2xy = 0
⇒ \(\frac{d y}{d x}=-\frac{2 x y}{x^2}=-\frac{2 y}{x}\)
Thus slope of tangent to curve (2) at (2, 3) = m2
= \(\left(\frac{d y}{d x}\right)_{(2,3)}\)
= – \(\frac{2 \times 3}{2}\) = – 3
Let θ be the acute angle between two given curves
∴ tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{-\frac{3}{2}+3}{1-\frac{3}{2} \times(-3)}\right|\)
⇒ tan θ = \(\left|\frac{\frac{3}{2}}{\frac{11}{2}}\right|\)
⇒ tan θ = \(\frac{3}{11}\)
⇒ θ = tan-1 (\(\frac{3}{11}\))
(ii) Given eqns. of curve be y = 4 – x2 ……….(1)
and y = x2 …………(2)
From (1) and (2) ; we get
x2 = 2
⇒ x = ± √2
From (1) ;
y = 4 – 2 = 2
Hence the point of intersection of two curves be (± √2, 2).
Diff. (1) w.r.t. x ; we get
\(\frac{d y}{d x}\) = – 2x …………….(3)
Diff. (2) w.r.t. x ; we get
\(\frac{d y}{d x}\) = 2x …………..(4)
at (√2, 2):
using (3) ;
\(\left(\frac{d y}{d x}\right)_{(\sqrt{2}, 2)}\) = m1 = – 2√2
using (4) ;
\(\left(\frac{d y}{d x}\right)_{(\sqrt{2}, 2)}\) = m2 = 2√2
Let θ be the angle of intersection of two curves
Then tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{-2 \sqrt{2}-2 \sqrt{2}}{1-8}\right|\)
= \(\frac{4 \sqrt{2}}{7}\)
∴ θ = tan-1 \(\frac{4 \sqrt{2}}{7}\)
at (- √2, 2):
using (3) ;
m1 = \(\left(\frac{d y}{d x}\right)_{(-\sqrt{2}, 2)}\) = 2√2
using (4) ;
m2 = \(\left(\frac{d y}{d x}\right)_{(-\sqrt{2}, 2)}\) = – 2√2
∴ tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{4 \sqrt{2}}{1-8}\right|\)
= \(\frac{4 \sqrt{2}}{7}\)
∴ θ = tan-1 (\(\frac{4 \sqrt{2}}{7}\))
(iii) given eqns. of curve be y2 = x ……….(1)
and x2 = y ………(2)
From (1) and (2) ; we have
x4 = x
⇒ x (x3 – 1) = 0
⇒ x = 0 and x3 – 1 = 0
i.e. (x – 1) (x2 + x + 1) = 0
⇒ x = 1 or x2 + x + 1 = 0
∴ only real values of x are 0 and 1.
When x = 0
∴ from (2) ; y = 0
When x = 1
∴ from (2) ; y = 1
Thus both curves intersects at (0, 0) and (1, 1)
Diff. both sides w.r.t. x, we get
2y \(\frac{d y}{d x}\) = 1
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2 y}\) ……….(3)
Diff. (2) both sides w.r.t. x ; we get
2x = \(\frac{d y}{d x}\) ……….(4)
at (0, 0):
∴ m1 = \(\left(\frac{d y}{d x}\right)_{(0,0)}\) = ∞
Thus tangent to curve (1) at (0, 0) is parallel to y-axis
and m2 = \(\left(\frac{d y}{d x}\right)_{(0,0)}\)
= 2 × 0 = 0
∴ tangent to curve (2) at (0, 0) is parallel to x axis.
Hence the angie between the tangents to given two curves at (0, 0) is right angle.
Hence given two curves intersects at right angle at (0, 0).
At (1, 1):
m1 = \(\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{1}{2}\) [using (3)]
and m2 = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = 2 [using (4)]
Let θ be the angle of intersection of two curves.
Then tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{\frac{1}{2}-2}{1+\frac{1}{2} \times 2}\right|=\left|\frac{3}{4}\right|\)
∴ θ = tan-1 \(\left(\frac{3}{4}\right)\).
Question 30.
Find the angle of intersection of the curves x2 + y = 4 and (x – 2)2 + y2 = 4 at the point ¡n tbe first quadrant.
Solution:
Given eqns. of curves are x2 + y2 = 4 ………….(1)
and (x – 2)2 + y2 = 4 …………..(2)
For point of intersection of both curves,
we solve eqn. (1) and eqn. (2) simultaneously.
eqn. (2) – eqn. (1) ; we have
(x – 2)2 – x2 = 0
⇒ x2 – 4x + 4 – x2 = 0
⇒ – 4x + 4 = 0
⇒ x = 1
∴ from (1) ;
y2 = 3
⇒ y = ± √3
So point of intersection of both curves are (1, ± √3) but we want to find the angle of intersection at the point which lies in first quadrant.
∴ point of intersection of both curves be P(1, √3).
Diff. eqn. (1) both sides w.r.t. x;
we have 2x + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{x}{y}\)
∴ slope of tangent to given curve (1) at P(1, √3) = m1
= \(\left(\frac{d y}{d x}\right)_{(1, \sqrt{3})}=-\frac{1}{\sqrt{3}}\)
Diff. eqn. (2) both sides w.r.t. x ; we have
2 (x – 2) + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{(x-2)}{y}\)
∴ slope of tangent to given curve (2) at P (1, √3) = m2
= \(\left(\frac{d y}{d x}\right)_{(1, \sqrt{3})}\)
= \(-\frac{(1-2)}{\sqrt{3}}=\frac{1}{\sqrt{3}}\)
Let θ be the acute angle between given curves
Then tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{-\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}}\right|\)
= \(\left|\frac{-\frac{2}{\sqrt{3}}}{\frac{2}{3}}\right|\) = √3
⇒ θ = \(\frac{\pi}{3}\)
Question 31.
Prove that the curves y = x2 – 3x + 1 and x (y + 3) = 4 intersect at right angles at the point (2, — 1).
Solution:
Eqn’s of given curve are;
y = x2 – 3x + 1 ………..(1)
and x (y + 3) = 4 ………..(2)
Diff. eqn. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = 2x – 3
∴ slope of tangent to given curve (1) at (2, – 1) = m1
= \(\left(\frac{d y}{d x}\right)_{(2,-1)}\)
= 2 2 – 3 = 1
Diff. eqn. (2) w.r.t. x, we have
\(\frac{d y}{d x}=-\frac{4}{x^2}\)
[∵ y = \(\frac{4}{x}\) – 3]
∴ slope of tangent to given curve (2) at (2, – 1) = m2
= \(\left(\frac{d y}{d x}\right)_{(2,-1)}\)
= – \(\frac{4}{2^2}\) = – 1
Here, m1m2 = 1 × (- 1) = – 1
Thus both curve sintersect at right angle at the point (2, – 1).
Question 32.
Find the condition that the curves 2x = y2 and 2xy = k intersect orthogonally. (NCERT Exemplar)
Solution:
Given curves are 2x = y2 ………..(1)
and 2xy = k ………….(2)
From (1) and (2) ;
y3 = k
⇒ y = k1/3
∴ from (1) ; we have
x = \(\frac{k^{2 / 3}}{2}\)
Thus both curves intersects at P (\(\frac{k^{2 / 3}}{2}\), k1/3)
Differentiate (1) ; w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{y}\)
∴ m1 = \(\left.\frac{d y}{d x}\right]_{a t \mathrm{P}}=\frac{1}{k^{1 / 3}}\)
Differentiate (2) ; w.r.t. x, we get
x \(\frac{d y}{d x}\) + y = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{y}{x}\)
∴ m2 = \(\left.\frac{d y}{d x}\right]_{\mathrm{at} P}\)
= \(\frac{-2 k^{1 / 3}}{k^{2 / 3}}=\frac{-2}{k^{1 / 3}}\)
Since the given curves cut right angles at P.
∴ m1 m2 = – 1
⇒ \(\left(\frac{1}{k^{1 / 3}}\right)\left(\frac{-2}{k^{1 / 3}}\right)\) = – 1
⇒ 2 = k2/3
⇒ k2 = 8.
Question 33.
Prove that the curves xy = 4 and x2 + y2 = 8 touch each other. (NCERT Exampler)
Solution:
Given eqns. of curve be xy = 4 ………..(1)
and x2 + y2 = 8 ………..(2)
From (2) ;
x2 + y2 – 2 × 4 = 0
⇒ x2 + y2 – 2xy = 0 [using (1)]
⇒ (x – y)2 = 0
⇒ x = y
∴ From (1) ;
x2 = 4
⇒ x = ± 2
∴ From (3) ;
⇒ y = ± 2
Hence the points of intersection of two curves be (± 2, ± 2).
Diff. both sides of eqn. (1) w.r.t. x ; we get
x \(\frac{d y}{d x}\) + y = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{y}{x}\) ………..(3)
Diff, both sides eqn. (2) w.r.t. x; we get
2x + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{x}{y}\)
at P (2, 2):
\(\left(\frac{d y}{d x}\right)_{c_1}=-\frac{2}{2}\) = – 1 and
\(\left(\frac{d y}{d x}\right)_{c_2}\) = – 1
∴ \(\left(\frac{d y}{d x}\right)_{c_1}=\left(\frac{d y}{d x}\right)_{c_2}\)
So two curves touch each other at p (2, 2).
Similarly both curves touch each other at other points (- 2, – 2); (- 2, 2) and (2, – 2).
Question 34.
Find the angle of Intersection between the curves y2 = 4x and x2 = 4y.
Solution:
Given equations of curves are
y2 = 4x …………..(1)
and x2 = 4y …………….(2)
from (2) ;
y = \(\frac{x^2}{4}\)
∴ from (1) ;
\(\left(\frac{x^2}{4}\right)^2\) = 4x
⇒ x4 = 64x
⇒ x (x3 – 64) = 0
⇒ x = 0, 4
∴ y = 0, 4
Thus points of intersection of two curves are Q (0, 0) and P(4, 4).
Diff. (1) and (2) w.r.t. x both sides ; we have
2y \(\frac{d y}{d x}\) = 4
⇒ \(\frac{d y}{d x}\) = \(\frac{2}{y}\)
and 2x = 4 \(\frac{d y}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{x}{2}\)
∴ m1 = slope of tangent to curve (1) at P
= \(\left(\frac{d y}{d x}\right)_{P(4,4)}\)
= \(\frac{2}{4}=\frac{1}{2}\)
m2 = slope of tangent to curve (2) at P (4, 4)
= \(\left(\frac{d y}{d x}\right)_{P(4,4)}\)
= \(\frac{4}{2}\) = 2
Let θ be the angle between the tangents to given curves then
tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{\frac{1}{2}-2}{1+\frac{1}{2} \times 2}\right|=\frac{3}{4}\)
⇒ θ = tan-1 \(\frac{3}{4}\)
and at point Q (0, 0) m1 → ∞ m2 → 0
(i.e. one tangent is y-axis and other tangent is x-axis)
Thus both tangents to given curves cuts orthogonally.