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KEAM 2013 Math Paper

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Question : 99 of 120

Marks: +1, -0

The slope of the normal to the curve

x=t^2+3t-8,

y=2t^2-2t-5

(2,-1)

is

$\frac{6}{7}$
$-\frac{6}{7}$
$\frac{7}{6}$
$-\frac{7}{6}$
$-\frac{1}{2}$

Solution:

Given curves are

x=t^{2}+3 t-8

\therefore \frac{dx}{dt}=2t+3

and

y=2t^{2}-2t-5

\therefore \frac{dy}{dt}=4t-2

Slope of tangent

=\frac{dy}{dx}=\frac{dy}{dt} \times \frac{dt}{dx}=\frac{4t-2}{2t+3}

......(i)

since, curve passes through the point (2,-1) .

\therefore t^{2}+3t-8=2

and

2t^{2}-2t-5=-1

\Rightarrow t^{2}+3t-10=0

and

2t^{2}-2t-4=0

\Rightarrow t^{2}+5t-2t-10=0

and

t^{2}-t-2=0

\Rightarrow (t+5)(t-2)=0

and

(t^{2}-2t+t-2)=0

\Rightarrow t=-5,2

and

(t-2)(t+1)=0

\Rightarrow t=-5,2

and

t=-1,2

So, common value of

t

is 2 .

On putting

t=2

in Eq. (i), we get

\left[\frac{dy}{dx}\right]_{\text{at } t=2}=\frac{4(2)-2}{2(2)+3}=\frac{6}{7}

Slope of normal

=\frac{\frac{-1}{dy}}{\frac{dy}{dx}}=\frac{-1}{\frac{6}{7}}=-\frac{7}{6}

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