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= 3xSolution

Slope of tangent
\[{\left( {\frac{{dy}}{{dx}}} \right)_{x = {x_1}}}\]

y= 3xdifferentiate 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 = 10Solution

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 = -1/64Question (3)

Find the slope of the tangent to curve y = xSolution

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 = xSolution

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 = 24Question (5)

Find the slope of the normal to the curve x=acosSolution

slope of normal =
\[ \text{slope of normal=} \frac{- 1}{{\frac{{dy}}{{dx}}}}\]

x = acosdifferentiate 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 = asin

Question (6)

Find the slope of the normal to the curve x=1-asinθ, y=bcosSolution

x = 1 - asinθdifferentiate with respect to θ \[\frac{{dx}}{{d\theta }} = - a\cos \theta \] y = bcos

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 = xSolution

y = xdifferentiate 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

3x

3(x

(x-3)(x-1) = 0

x-3= 0 OR x+1 = 0

x = 3 OR x = -1

If x=3, y = x

y= 27 - 27 - 27 +7 = -20

(x, y) = (3, -20)

If x = -1 , y = x

y = -1 -3 +9 +7 = 12

(x, y) = (-1, 12)

Question (8)

Find a point on the curve y=(x-2)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)

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)

∴ (x, y) = (3, 1)

Question (9)

Find the point on the curve y=xSolution

If y = mx + c is equation of line

slope of line = m

Equation of tangent = y = x - 11slope of line = m

Slope of tangent = 1 ---(1)

y = x

differentiate with respect to x

\[\frac{{dy}}{{dx}} = 3{x^2} - 11\] Slope of tangent = 3x

From (1) and (2)

3x

3x

x

x = ±2 If x=2, y=x

(x, y) = (2, -9)

If x=-2, y=x

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 (x_{1}, y_{1}) with slope m is given by y-y_{1} = m(x-x_{1})

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)

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)

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

\[ \Rightarrow \frac{{{x^2}}}{9} = 1\] x

⇒ 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 = x

(ii) y = x

(iii) y = x

(iv) y = x

Solution

(i) y = xdifferentiate 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 = x

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 = x

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 = x

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=x(a) parallel to the line 2x - y + 9 = 0

(b) perpendicular to the line 5y-15x = 13

Solution

y = xdifferentiate 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)

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 = 7xSolution

y = 7xdifferentiate 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=xSolution

y = xdifferentiate with respect to x

\[\frac{{dy}}{{dx}} = 3{x^2}\] slope of tangent = y-coordinate

3x

x

⇒ x

If x = 0, y = 0

If x = 3, y = 27

points are (0, 0) and (3, 27)

Question (18)

For the curve y=4xSolution

y=4x\[\frac{{dy}}{{dx}} = 12{x^2} - 10{x^4}\] Let the tangent be drawn at (x

∴ (x

y

\[\frac{{dy}}{{dx}} = 12x_0^2 - 10x_0^4\] The equation of tangent is given by

y - y

y - y

it passes through (0, 0)

∴ 0 - y

- y

y

substituting y

4x

-8x

8x

x

x

x

If x

y

If x

y

If x

y

So points are (0, 0), (1, 2) and (-1, -2)

Question (19)

Find the points on the curve xSolution

y = xdifferentiate 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+y

y = ±2

so points are (1, 2) and (1, -2)

Question (20)

Find the equation of the normal at the point (amSolution

aydifferentiate 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 - 3am

2x+3my = 2am

Question (21)

Find the equation of the normals to the curve y = xSolution

line : x +14y+4 = 0 \[\text{slope of line =} \frac{{ - a}}{b} = \frac{{ - 1}}{{14}}\] y = xdifferentiate with respect to x

\[\frac{{dy}}{{dx}} = 3{x^2} + 2\] slope of tangent = 3x

\[ \text{slope of normal =} \frac{{ - 1}}{{3{x^2} + 2}}\] normal is parallel to line

slope of normal = slope of line \[\therefore \frac{{ - 1}}{{3{x^2} + 2}} = \frac{{ - 1}}{{14}}\] 3x

3x

x

⇒x ±2

IF x=2, y = (2)

(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)

(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 ySolution

ydifferentiate 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 - 2at

x-ty+at

\[\text{slope of normal =} \frac{{ - 1}}{{\text{slope of tangent}}}\] ∴ slope of normal = -t

Equation of normal is

y - 2at = -t (x-at

y-2at = -xt+at

tx+y = at

Question (23)

Prove that the curves x=ySolution

x=yReplacing value x from (1) in equation(2)

y

y

y = k

given 8k

k

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=y

\[y = {\left( {{2^{\frac{{ - 1}}{2}}}} \right)^2} = {2^{ - 1}} = \frac{1}{2}\] \[\left( {x,y} \right) = \left( {\frac{1}{2},\frac{1}{{\sqrt 2 }}} \right)\] Now y

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 m

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 m

Now m

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 (xSolution

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 - y

Question (25)

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

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

Question (26)

The slope of the normal to the curve y=2x(A) 3 (B) 1/3

(C) -3 (D) -1/3

Solution

y=2xdifferentiate 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 y(A) (1, 3) (B) (2, 1)

(C) (1, -2) (D) (-1, 2)

Solution

y = x + 1 is tangentslope of tangent m = 1 ---(1)

y

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

y

x = 1

so (x, y) = (1, 2) So option (A) is correct