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KEAM 2010 Math Paper

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Question : 38 of 120

Marks: +1, -0

The angle between the curves,

y=x^{2}

y^{2}-x=0

(1,1),

is

$\;\frac{\pi}{2}$
$\tan^{-1}\;\frac{4}{3}$
$\;\frac{\pi}{3}$
$\;\frac{\pi}{4}$
$\tan^{-1}\;\frac{3}{4}$

Solution:

Given curve are

y=x^{2}

....(i)

and

y^{2}-x=0

....(ii)

From Eq (i),

\;\frac{dy}{dx}=2x

\therefore\;\left(\;\frac{dy}{dx}\right)_{(1,1)}=2=m_{1}

(say)

From Eq. (ii),

2y\;\frac{dy}{dx}-1=0

\Rightarrow\;\frac{dy}{dx}=\frac{1}{2y}

\therefore\;\left(\frac{dy}{dx}\right)_{(1,1)}=\;\frac{1}{2}=m_{2}

(say)

Let

\theta

be the angle between given curve. Then,

\tan\theta =\;\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}

=\;\frac{2-\frac{1}{2}}{1+2\cdot\;\frac{1}{2}} =\;\frac{\frac{3}{2}}{2} =\;\frac{3}{4}

\therefore\;\theta =\tan^{-1}\;\frac{3}{4}

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