12th NCERT Application of Derivatives Exercise 6.3 Questions 27
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Question (1)

Find the slope of the tangent of the tangent to the curve y= 3x4 - 4x at x=4

Solution

Slope of tangent \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = {x_1}}}\]
y= 3x4 - 4x
differentiate with respect to x \[\frac{{dy}}{{dx}} = 12{x^3} - 4\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 4}} = 12{\left( 4 \right)^3} - 4\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 4}} = 12 \times 64 - 4\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 4}} = 768 - 4 = 764\]

Question (2)

Find the slope of the tangent of the tangent to the curve \[y = \frac{{x - 1}}{{x - 2}}\] x ≠ 2 at x = 10

Solution

differentiate with respect to x \[\frac{{dy}}{{dx}} = \frac{{\left( {x - 2} \right)\frac{d}{{dx}}\left( {x - 1} \right) - \left( {x - 1} \right)\frac{d}{{dx}}\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{\left( {x - 2} \right)\left( 1 \right) - \left( {x - 1} \right)\left( 1 \right)}}{{{{\left( {x - 2} \right)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{x - 2 - x + 1}}{{{{\left( {x - 2} \right)}^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{\left( {x - 2} \right)}^2}}}\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 10}} = \frac{{ - 1}}{{{{\left( {10 - 2} \right)}^2}}} = \frac{{ - 1}}{{64}}\] slope of tangent = -1/64

Question (3)

Find the slope of the tangent to curve y = x3 - x + 1 at the point whose x-coordinate is 2

Solution

differentiate with respect to x \[\frac{{dy}}{{dx}} = 3{x^2} - 1\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 2}} = 3{\left( 2 \right)^2} - 1\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 2}} = 12 - 1 = 11\]

Question (4)

Find the slope of the tangent to curve y = x3 - 3x + 2 at the point whose x-coordinate is 3

Solution

differentiate with respect to x \[\frac{{dy}}{{dx}} = 3{x^2} - 3\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 3}} = 3{\left( 3 \right)^2} - 3\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 3}} = 27 - 3 = 24\] slope of tangent = 24

Question (5)

Find the slope of the normal to the curve x=acos3θ, y=a sin3θ at θ = π/4

Solution

slope of normal = \[ \text{slope of normal=} \frac{- 1}{{\frac{{dy}}{{dx}}}}\]
x = acos3θ
differentiate with respect to θ \[\frac{{dx}}{{d\theta }} = a \times 3{\cos ^2}\theta \left( { - \sin \theta } \right)\] \[\frac{{dx}}{{d\theta }} = - 3a{\cos ^2}\theta \sin \theta \] y = asin3θ differentiate with respect to θ \[\frac{{dy}}{{d\theta }} = a \cdot 3{\sin ^2}\theta \left( {\cos \theta } \right)\] \[\frac{{dy}}{{d\theta }} = 3a{\sin ^2}\theta \left( {\cos \theta } \right)\] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}\] \[\frac{{dy}}{{dx}} = \frac{{3a{{\sin }^2}\theta \cos \theta }}{{ - 3a{{\cos }^2}\theta \sin \theta }}\] \[\frac{{dy}}{{dx}} = - \tan \theta \] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\theta = \frac{\pi }{4}}} = - \tan \frac{\pi }{4} = - 1\] slope of normal = \[{\text{slope of normal = }}\frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\theta = \frac{\pi }{4}}}}} = \frac{{ - 1}}{{ - 1}} = 1\]

Question (6)

Find the slope of the normal to the curve x=1-asinθ, y=bcos2θ at θ=π/2

Solution

x = 1 - asinθ
differentiate with respect to θ \[\frac{{dx}}{{d\theta }} = - a\cos \theta \] y = bcos2θ
differentiate with respect to θ \[\frac{{dy}}{{d\theta }} = b2\cos \theta \left( { - \sin \theta } \right)\] \[\frac{{dy}}{{d\theta }} = - 2b\sin \theta \cos \theta \] \[\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}\] \[\frac{{dy}}{{dx}} = \frac{{ - 2b\sin \theta \cos \theta }}{{ - a\cos \theta }}\] \[\frac{{dy}}{{dx}} = \frac{{2b}}{a}\sin \theta \] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\theta = \frac{\pi }{2}}} = \frac{{2b}}{a}\sin \frac{\pi }{2} = \frac{{2b}}{a}\] \[\text{slope of normal=} \frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\theta = \frac{\pi }{2}}}}}\] \[\text{slope of normal=} \frac{{ - 1}}{{\frac{{2b}}{a}}}\] \[\text{slope of normal=} \frac{{ - a}}{{2b}}\]

Question (7)

Find points at which the tangent to the curve y = x3 -3x2 - 9x + 7 is parallel to the x-axis

Solution

y = x3 -3x2 - 9x + 7
differentiate with respect to x
slope of tangent = dy/dx \[\frac{{dy}}{{dx}} = 3{x^2} - 6x - 9\] Slope of x-axis = 0
tangent parallel to x-axis
slope of tangent = slope of x-axis
3x2 -6x - 9 = 0
3(x2 - 2x -3) =0
(x-3)(x-1) = 0
x-3= 0 OR x+1 = 0
x = 3 OR x = -1
If x=3, y = x3 -3x2 - 9x + 7
y= 27 - 27 - 27 +7 = -20
(x, y) = (3, -20)
If x = -1 , y = x3 -3x2 - 9x + 7
y = -1 -3 +9 +7 = 12
(x, y) = (-1, 12)

Question (8)

Find a point on the curve y=(x-2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4)

Solution

tangent parallel to AB, A(2, 0) and B(4, 4)
\[\text{slope of}\quad \overline {AB} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\] \[\text{slope of}\quad \overline {AB} = \frac{{4 - 0}}{{4 - 2}} = 2\] y = (x - 2)2
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 2\left( {x - 2} \right)\] As tangent parallel AB
slope of tangent = slope of AB
2(x-2) = 2
x - 2 = 1
x = 3
If x=3, y=(x-2)2 = (3-2)2 = 1
∴ (x, y) = (3, 1)

Question (9)

Find the point on the curve y=x2 - 11x + 5 at which the tangent is y = x - 11

Solution

If y = mx + c is equation of line
slope of line = m
Equation of tangent = y = x - 11
Slope of tangent = 1 ---(1)
y = x3 - 11x + 5
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 3{x^2} - 11\] Slope of tangent = 3x2 - 11 ---(2)
From (1) and (2)
3x2 - 11 = 1
3x2 = 12
x2 = 4
x = ±2 If x=2, y=x3 -11x +5 then, y=23 -11(2) +5 =-9
(x, y) = (2, -9)
If x=-2, y=x3 -11x +5 then, y=(-2)3 -11(-2) +5 = 19 (x, y) = (-2, 19)
But (-2, 19) does not satisfy the equation
So point on curve is (2, -9) of tangent

Question (10)

Find the equation of all lines having slope -1 that are tangents to the curve \[y = \frac{1}{{x - 1}},x \ne 1\]

Solution

Equation of line passing through (x1, y1) with slope m is given by y-y1 = m(x-x1)
Given slope = -1 ---(1) and equation \[y = \frac{1}{{x - 1}},x \ne 1\] differentiate with respect to x
\[\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}}\] \[\text{slope of tangent=}\frac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}}---(2)\] From (1) and (2) \[\frac{{ - 1}}{{{{\left( {x - 1} \right)}^2}}} = - 1\] (x-1)2 = 1
x-1 = ± 1
⇒ x - 1 = 1 OR x-1 = -1
x = 2 or x =0
If x = 2, y = 1/(2-1) = 1
(x, y) = (2, 1)
If x = 0, y = 1/(0-1) = -1
(x, y) = (0, -1)
So equation of tangent at (2, 1) is
y - 1 = -1(x-2)
y - 1 = -x + 2
x + y - 3 = 0
Equation of tangent ay (0, -1) will be
y + 1 = -1( x - 0)
y + 1 = -x
x + y + 1 = 0
So equation of tangents are
x+y-3=0 and x+y+1=0

Question (11)

Find the equation of all lines having slope 2 which are tangents to the curve \[y = \frac{1}{{x - 3}},x \ne 3\]

Solution

Slope of tangent = 2 ---(1) \[y = \frac{1}{{x - 3}},x \ne 3\] differentiate with respect to x
\[\text{slope of tangent=}\frac{{ - 1}}{{{{\left( {x - 3} \right)}^2}}}\] From (1) and (2)
\[\frac{{ - 1}}{{{{\left( {x - 3} \right)}^2}}} = 2\] (x - 3) 2 = -½ Which is not possible as square can not be negative
So, no tangent can be drawn having slope 2

Question (12)

Find the equations of all lines having slope 0 which are tangent to the curve \[y = \frac{1}{{{x^2} - 2x + 3}}\]

Solution

Slope of tangent = 0 --- (1) \[y = \frac{1}{{{x^2} - 2x + 3}}\] differentiate with respect to x
\[\frac{{dy}}{{dx}} = \frac{{ - 1}}{{{{\left( {{x^2} - 2x + 3} \right)}^2}}} \times \left( {2x - 2} \right)\] \[\frac{{dy}}{{dx}} = \frac{{ - \left( {2x - 2} \right)}}{{{{\left( {{x^2} - 2x + 3} \right)}^2}}} - - - (2)\] From equation (1) and (2) \[\frac{{ - \left( {2x - 2} \right)}}{{{{\left( {{x^2} - 2x + 3} \right)}^2}}} = 0\] ⇒ 2x-2 = 0
x = 1
If x=1 \[y = \frac{1}{{{{\left( 1 \right)}^2} - 2\left( 1 \right) + 3}} = \frac{1}{2}\] So the tangent passes through (x, y) = (1, ½) Equation of tangent will be
y-½ = 0(x-1)
2y - 1 = 0

Question (13)

Find points on the curve \[\frac{{{x^2}}}{9} + \frac{{{y^2}}}{{16}} = 1\] at which the tangents are
(i) parallel to x-axis
(ii) parallel to y-axis

Solution

differentiate with respect to x
\[\frac{{2x}}{9} + \frac{{2y}}{{16}}\frac{{dy}}{{dx}} = 0\] \[\frac{x}{9} + \frac{y}{{16}}\frac{{dy}}{{dx}} = 0\] \[\frac{{dy}}{{dx}} = \frac{{ - 16}}{9}\frac{x}{y}\] \[\frac{{dy}}{{dx}} = \frac{{ - 16x}}{{9y}}\] \[\text{slope of tangent=}\frac{{ - 16x}}{{9y}}\] (a) tangent parallel to x-axis
slope of x-axis = 0
Slope of tangent = slope of x-axis
\[\frac{{ - 16x}}{{9y}} = 0\] ⇒ x = 0
If x =0 ⇒ \[ \Rightarrow \frac{{{y^2}}}{{16}} = 1\] y = ±4 Points of tangents are (0, 4), (0, -4)
(ii) Parallel to y-axis
Slope of y-axis is not defined that is denominator = 0 ∴ 9y = 0
y = 0 If y=0
\[ \Rightarrow \frac{{{x^2}}}{9} = 1\] x2 = 9
⇒ x = ±3 Points through which tangents parallel to y-axis are drawn are (3,0)and (-3, 0)

Question (14)

Find the equations of tangent and normal to the given curves at the indicated points
(i) y = x4 -6x3+13x2 -10x + 5 at (0, 5)
(ii) y = x4 -6x3+13 x2 -10x + 5 at (1, 3)
(iii) y = x3 at (1, 1)
(iv) y = x2 at (0, 0) (v) x = cost, y=sin t at t = π/4

Solution

(i) y = x4 -6x3+13 x2- 10x +5
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 4{x^3} - 18{x^2} + 26x - 10\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 0}} = - 10\] slope of tangent = -10
Equation of tangent through (0, 5) is
y - 5 = -10(x-0)
10x + y - 5=0 \[ \text{slope of normal=} \frac{{ - 1}}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{x = 0}}}}\] Equation of normal \[y - 5 = \frac{1}{{10}}\left( {x - 0} \right)\] 10y - 50 = x
x - 10y + 50 = 0

(ii) y = x4 -6x3+13 x2 -10x + 5
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 4{x^3} - 18{x^2} + 26x - 10\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 1}} = 4{\left( 1 \right)^3} - 18{\left( 1 \right)^2} + 26\left( 1 \right) - 10\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 1}} = 4 - 18 + 26 - 10\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 1}} = 2\] ∴ slope of tangent = 2
Equation of tangent through (1, 3) is
y-3 = 2(x-1)
2x - y + 1 =0
\[\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}\] \[\text{slope of normal =} \frac{{ - 1}}{2}\] Equation of normal is
\[y - 3 = \frac{{ - 1}}{2}\left( {x - 1} \right)\] 2y - 6 = -x + 1
x + 2y = 7

(iii) y = x3
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 3{x^2}\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,1} \right)}} = 3\] slope of tangent = 3
Equation of tangent is
y-1 = 3(x - 1)
3x - y - 2 = 0
\[\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}\] \[\text{slope of normal =} \frac{{ - 1}}{3}\] Equation of normal is
\[y - 1 = \frac{{ - 1}}{3}\left( {x - 1} \right)\] 3y -3 = -x + 1
x + 3y = 4

(iv) y = x2
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 2x\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {0,0} \right)}} = 0\] slope of tangent is = 0
Equation of tangent is y - 0 = 0(x - 0)
y=0
\[\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}\] \[\text{slope of normal =} \frac{{ - 1}}{0}\] slope of normal is not defined
so normal is parallel to y-axis and passes through (0, 0) it is y-axis so equation is x = 0
(v) x = cost
differentiate with respect to x
\[\frac{{dx}}{{dt}} = - \sin t\] y = sint
differentiate with respect to x
\[\frac{{dy}}{{dt}} = \cos t\] \[\text{Now}\quad \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\] \[\frac{{dy}}{{dx}} = \frac{{\cos t}}{{ - \sin t}} = - \cot t\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{t = \frac{\pi }{4}}} = - \cot \frac{\pi }{4} = - 1\] slope of tangent = -1
Equation of tangent is \[y - \frac{1}{{\sqrt 2 }} = - 1\left( {x - \frac{1}{2}} \right)\] \[\sqrt 2 y - 1 = - \sqrt 2 x + 1\] \[\sqrt 2 y + \sqrt 2 x = 2\] \[y + x = \sqrt 2 \] slope of normal = -1/-1 = 1
Equation of normal is \[y - \frac{1}{{\sqrt 2 }} = \left( {x - \frac{1}{{\sqrt 2 }}} \right)\] x - y = 0

Question (15)

Find the equation of the tangent line to the curve y=x2 - 2x + 7 which is
(a) parallel to the line 2x - y + 9 = 0
(b) perpendicular to the line 5y-15x = 13

Solution

y = x2 - 2x + 7
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 2x - 2\] slope of tangent = 2x - 2
(a) parallel to 2x - y + 9 = 0
Equation of line is 2x - y + 9 = 0
slope of line = -a/b = -2/-1 = 2
tangent parallel line
∴ slope of tangent = slope of line
∴ 2x - 2 = 2
2x = 4
x = 2
If x=2, y = (2)2 - 2(2) + 7 = 7
Equation of tangent is
y - 7 = 2(x-2)
2x - y + 3 = 0
(b) Perpendicular to 5y -15x = 13
-15x + 5y = 13
slope of line = -a/b = -15/5 = 3
tangent is perpendicular to line
∴ slope of tangent × slope of line = -1
∴ (2x-2) × 3 = -1
6x - 6 = -1
6x = 5
x = 5/6
If x = 5/6,
\[y = {\left( {\frac{5}{6}} \right)^2} - 2\left( {\frac{5}{6}} \right) + 7\] \[y = \frac{{25}}{{36}} - \frac{{10}}{6} + 7\] \[y = \frac{{25 - 60 + 252}}{{36}} = \frac{{217}}{{36}}\] Equation of tangent is \[y - \frac{{217}}{{36}} = \frac{{ - 1}}{3}\left( {x - \frac{5}{6}} \right)\] 36y - 217 = -12x + 10
12x + 36y - 227 = 0

Question (16)

Show that the tangents to the curve y = 7x3 + 11 at the points where x=2 and x=-2 are parallel

Solution

y = 7x3 + 11
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 21{x^2}\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 2}} = 21{\left( 2 \right)^2} = 84\] slope of tangent at x=2 is 84
\[{\left( {\frac{{dy}}{{dx}}} \right)_{x = - 2}} = 21{\left( { - 2} \right)^2} = 84\] slope of tangent at x=-2 is 84
slope are equal so tangents are parallel

Question (17)

Find the points on the curve y=x3 at which the slope of the tangent is equal to the y-coordinate of the point

Solution

y = x3
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 3{x^2}\] slope of tangent = y-coordinate
3x2 = x3
x3 - 3x2 = 0
⇒ x2 ( x - 3) = 0
If x = 0, y = 0
If x = 3, y = 27
points are (0, 0) and (3, 27)

Question (18)

For the curve y=4x3 - 2x5, find all the points at which the tangent passes through the origin

Solution

y=4x3 - 2x5 differentiate with respect to x
\[\frac{{dy}}{{dx}} = 12{x^2} - 10{x^4}\] Let the tangent be drawn at (x0, y0)
∴ (x0, y0) ∈ curve
y0 = 4x03 - 2x05 ---(1)
\[\frac{{dy}}{{dx}} = 12x_0^2 - 10x_0^4\] The equation of tangent is given by
y - y0 = m(x - x0)
y - y0 = (12x02 - 10x04)(x - x0)
it passes through (0, 0)
∴ 0 - y0 = (12x02 - 10x04)(0 - x0)
- y0 = - x0(12x02 - 10x04)
y0 = 12x03 - 10x05
substituting y0 from (1)
4x03 - 2x05 = 12x03 - 10x05
-8x03 + 8x05 = 0
8x03( x02 - 1) =0
x03( x0 - 1)( x0 + 1) = 0
x0 = 0 OR x0 - 1 = 0 OR x0 + 1 = 0
x0 = 0 OR x0 = 1 OR x0 = - 1
If x0 = 0
y0 = 4x03 - 2x05 =0 ⇒ (0, 0)
If x0 = 1
y0 = 4(1)3 - 2(1)5 =2 ⇒ (1, 2)
If x0 = -1
y0 = 4(-1)3 - 2(-1)5 = -2 ⇒ (-1, -1)
So points are (0, 0), (1, 2) and (-1, -2)

Question (19)

Find the points on the curve x2 + y2 - 2x - 3 = 0 at which the tangents are parallel to the x-axis

Solution

y = x2 + y2 - 2x - 3 = 0
differentiate with respect to x
\[2x + 2y\frac{{dy}}{{dt}} - 2 = 0\] \[y\frac{{dy}}{{dt}} = 1 - x\] \[\frac{{dy}}{{dt}} = \frac{{1 - x}}{y}\] slope of tangent at (x, y) = (1-x)/y
slope of x-axis = 0
tangent parallel to x-axis
slope of tangent = slope of x-aaxis
\[\therefore \frac{{1 - x}}{y} = 0\] x= 1
If x=1,
1+y2-2-3 = 0 ⇒ y2 = 4,
y = ±2
so points are (1, 2) and (1, -2)

Question (20)

Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3

Solution

ay2 = x3
differentiate with respect to x
\[2ay\frac{{dy}}{{dx}} = 3{x^2}\] \[\frac{{dy}}{{dx}} = \frac{{3{x^2}}}{{2ay}}\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {a{m^2},a{m^3}} \right)}} = \frac{{3{{\left( {a{m^2}} \right)}^2}}}{{2a\left( {a{m^3}} \right)}}\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {a{m^2},a{m^3}} \right)}} = \frac{{3{a^2}{m^4}}}{{2{a^2}{m^3}}}\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {a{m^2},a{m^3}} \right)}} = \frac{3}{2}m\] slope of tangent = 3m/2
\[\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}\] \[ \text{slope of normal =} \frac{{ - 2}}{{3m}}\] Equation of normal is \[y - a{m^3} = \frac{{ - 2}}{{3m}}\left( {x - a{m^2}} \right)\] 3my - 3am4 = -2x +2am2
2x+3my = 2am2 + 3am4

Question (21)

Find the equation of the normals to the curve y = x3 + 2x +6 which are parallel to the line x +14y+4 = 0

Solution

line : x +14y+4 = 0 \[\text{slope of line =} \frac{{ - a}}{b} = \frac{{ - 1}}{{14}}\] y = x3 + 2x +6
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 3{x^2} + 2\] slope of tangent = 3x2 + 2
\[ \text{slope of normal =} \frac{{ - 1}}{{3{x^2} + 2}}\] normal is parallel to line
slope of normal = slope of line \[\therefore \frac{{ - 1}}{{3{x^2} + 2}} = \frac{{ - 1}}{{14}}\] 3x2 + 2 = 14
3x2 = 12
x2 = 4
⇒x ±2
IF x=2, y = (2)3+2(2) + 6 = 18
(x, y ) = (2, 18)
Equation of normal through (2, 18) is
\[y - 18 = \frac{{ - 1}}{{14}}\left( {x - 2} \right)\] 14y -252 = -x +2
x+14y-254= 0
If x = -2, y = (-2)3+2(-2) + 6 = -6
(x, y) = (-2, -6)
Equation of normal through (-2, -6) is given by
\[y + 6 = \frac{{ - 1}}{{14}}\left( {x + 2} \right)\] 14y +84=-x-2
x+14y+86=0

Question (22)

Find the equations of the tangent and normal to the parabola y2 = 4ax at the point (at2, 2at)

Solution

y2 = 4ax
differentiate with respect to x
\[2y\frac{{dy}}{{dx}} = 4a\] \[\frac{{dy}}{{dx}} = \frac{{4a}}{{2y}} = \frac{{2a}}{y}\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {a{t^2},2at} \right)}} = \frac{{2a}}{{2at}} = \frac{1}{t}\] slope of tangent = 1/t
Equation of tangent is
\[y - 2at = \frac{1}{t}\left( {x - a{t^2}} \right)\] ty - 2at2 = x-at2
x-ty+at2 = 0
\[\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}\] ∴ slope of normal = -t
Equation of normal is
y - 2at = -t (x-at2)
y-2at = -xt+at3
tx+y = at3 +2at

Question (23)

Prove that the curves x=y2 and xy =k cut at right angles if 8k2 = 1

Solution

x=y2 ---(1) and xy =k ---(2)
Replacing value x from (1) in equation(2)
y2y = k
y3 = k
y = k1/3 ---(3)
given 8k2 =1
k2 = 1/8
k = 1/√8
Replacing value of k in equation (3) \[y = {\left( {\frac{1}{{\sqrt 8 }}} \right)^{\frac{1}{3}}} = {\left( {{2^{\frac{{ - 3}}{2}}}} \right)^{\frac{1}{3}}} = {2^{\frac{{ - 1}}{2}}}\] x=y2
\[y = {\left( {{2^{\frac{{ - 1}}{2}}}} \right)^2} = {2^{ - 1}} = \frac{1}{2}\] \[\left( {x,y} \right) = \left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)\] Now y2 = x
differentiate with respect to x
\[2y\frac{{dy}}{{dx}} = 1\] \[\frac{{dy}}{{dx}} = \frac{1}{{2y}}\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)}} = \frac{1}{{2 \times \frac{1}{{\sqrt 2 }}}} = \frac{1}{{\sqrt 2 }}\] ∴ slope of tangent to first curve m1 = 1/√2
Now xy = k
differentiate with respect to x
\[x\frac{{dy}}{{dx}} + y = 0\] \[\frac{{dy}}{{dx}} = \frac{{ - y}}{x}\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)}} = \frac{{ - \frac{1}{{\sqrt 2 }}}}{{\frac{1}{2}}} = - \frac{1}{{\sqrt 2 }} \times 2\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)}} = - \sqrt 2 \] ∴ slope of tangent to second curve m2 = - √2
Now m1 m2 = 1/√2×- √2 = -1
slope of tangents are perpendicular to each other
so curve are orthogonal

Question (24)

Find the equation of the tangent and normal to the hyperbola \[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\] at the point (x0, y0)

Solution

differentiate with respect to x
\[\frac{{2x}}{{{a^2}}} - \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0\] \[\frac{y}{{{b^2}}}\frac{{dy}}{{dx}} = \frac{x}{{{a^2}}}\] \[\frac{{dy}}{{dx}} = \frac{{\frac{x}{{{a^2}}}}}{{\frac{y}{{{b^2}}}}} = \frac{{{b^2}x}}{{{a^2}y}}\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{\left( {{x_0},{y_{}}} \right)}} = \frac{{{b^2}{x_0}}}{{{a^2}{y_0}}}\] \[\text{slope of tangent =} \frac{{{b^2}{x_0}}}{{{a^2}{y_0}}}\] Equation of tangent is
y - y0 = m(x -x0) \[y - {y_0} = \frac{{{b^2}{x_0}}}{{{a^2}{y_0}}}\left( {x - {x_0}} \right)\] \[\frac{{y{y_0} - y_0^2}}{{{b^2}}} = \frac{{{x_0}x - x_0^2}}{{{a^2}}}\] \[\frac{{y{y_0}}}{{{b^2}}} - \frac{{y_0^2}}{{{b^2}}} = \frac{{{x_0}x}}{{{a^2}}} - \frac{{x_0^2}}{{{a^2}}}\] \[\frac{{{x_0}x}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} = \frac{{x_0^2}}{{{a^2}}} - \frac{{y_0^2}}{{{b^2}}}\] \[\frac{{{x_0}x}}{{{a^2}}} - \frac{{y{y_0}}}{{{b^2}}} = 1\] \[\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}\] \[\text{slope of normal =} - \frac{{{a^2}{y_0}}}{{{b^2}{x_0}}}\] Equation of normal \[y - {y_0} = - \frac{{{a^2}{y_0}}}{{{b^2}{x_0}}}\left( {x - {x_0}} \right)\] \[\frac{{y - {y_0}}}{{{a^2}{y_0}}} = - \frac{{\left( {x - {x_0}} \right)}}{{{b^2}{x_0}}}\] \[\therefore \frac{{\left( {x - {x_0}} \right)}}{{{b^2}{x_0}}} + \frac{{y - {y_0}}}{{{a^2}{y_0}}} = 0\]

Question (25)

Find the equation of the tangent to the curve y = √(3x-2) which is parallel to the line 4x-2y+5 = 0

Solution

line l: 4x-2y+5 =0 slope of line = -a/b = -4/-2 = 2
\[y = \sqrt {3x - 2} \] differentiate with respect to x
\[\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {3x - 2} }} \times 3\] \[\frac{{dy}}{{dx}} = \frac{3}{{2\sqrt {3x - 2} }}\] tangent parallel to line
slope of tangent = slope of line \[\frac{3}{{2\sqrt {3x - 2} }} = 2\] \[3 = 4\sqrt {3x - 2} \] 9 = 16(3x-2)
9 = 48x - 32
48x = 41
x = 41/48 \[\text{If x =} \frac{{41}}{{48}}\] \[y = \sqrt {3x - 2} \] \[y = \sqrt {3\left( {\frac{{41}}{{48}}} \right) - 2} \] \[y = \sqrt {\frac{{41 - 32}}{{16}}} = \sqrt {\frac{9}{{16}}} = \frac{3}{4}\] \[pt\left( {x,y} \right) = \left( {\frac{{41}}{{48}},\frac{3}{4}} \right)\] Equation of tangent is \[y - \frac{3}{4} = 2\left( {x - \frac{{41}}{{48}}} \right)\] \[\frac{{4y - 3}}{4} = \frac{{96x - 82}}{{48}}\] 48y - 36 = 96x-82
96x - 48y - 46 =0
48x - 24y - 23 = 0

Choose the correct answer in Exercises 26 and 27

Question (26)

The slope of the normal to the curve y=2x2 +3sinx at x= 0 is
(A) 3     (B) 1/3
(C) -3     (D) -1/3

Solution

y=2x2 +3sinx
differentiate with respect to x
\[\frac{{dy}}{{dx}} = 4x + 3\cos x\] \[{\left( {\frac{{dy}}{{dx}}} \right)_{x = 0}} = 0 + 3\cos 0 = 3\] Slope of tangent = 3 \[\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}\] \[\text{slope of normal =} \frac{{ - 1}}{3}\] So option (D) is correct

Question (27)

The line y = x + 1 is a tangent to the curve y2 = 4x at the point
(A) (1, 3)     (B) (2, 1)
(C) (1, -2)     (D) (-1, 2)

Solution

y = x + 1 is tangent
slope of tangent m = 1 ---(1)
y2 = 4x
differentiate with respect to x
\[2y\frac{{dy}}{{dx}} = 4\] \[y\frac{{dy}}{{dx}} = 2\] \[\frac{{dy}}{{dx}} = \frac{2}{y}\] \[ \text{slope of tangent =} \frac{2}{y}---(2)\] \[\therefore \frac{2}{y} = 1\] y = 2
y2 = 4x
x = 1
so (x, y) = (1, 2) So option (A) is correct
Exercise 6.2⇐
⇒ Exercise 6.4