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EAMCET Engineering 2008 Solved Paper

Section: Mathematics

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Question : 147 of 160

Marks: +1, -0

The angle between the curves

y^2=4 x+4

y^2=36(9-x)

$30^{\circ}$
$45^{\circ}$
$60^{\circ}$
$90^{\circ}$

Solution:

Given curves are

y^2=4 x+4 \;\text{ and }\; y^2=36(9-x)

. . . (i)

On solving, we get the points

(8,6)

and

(8,-6)

On differentiating Eq. (i), we get

\;2 y \;\frac{dy}{dx}=4 \;\; \;\text{ and }\; \;\; 2 y \;\frac{dy}{dx}=-36

\Rightarrow \;\; \;\frac{dy}{dx}=\;\frac{2}{y} \;\; \;\text{ and }\; \;\; \;\frac{dy}{dx}=\;\frac{-18}{y}

At point

(8,6)

,

m_1=\;\frac{dy}{dx}=\;\frac{2}{6}=\;\frac{1}{3}

and

m_2=\;\frac{dy}{dx}=\;\frac{-18}{6}=-3

\therefore \;\; \tan \theta=\;\frac{\;\frac{1}{3}+3}{1+\;\frac{1}{3} \times(-3)}=\;\frac{\frac{10}{3}}{0}=\infty

\Rightarrow \;\; \theta=\;\frac{\pi}{2}

And at point

(8,-6)

,

\;m_1=\;\frac{dy}{dx}=\;\frac{2}{-6}=-\;\frac{1}{3} \;\text{ and }\; m_2=\;\frac{dy}{dx}=\;\frac{-18}{-6}=3

\; \therefore \;\; \tan \theta=\;\frac{\left| -\frac{1}{3}-3 \right|}{1+ \left(-\;\frac{1}{3}\right) \cdot 3}=\;\frac{\frac{10}{3}}{0}=\infty

\;\Rightarrow \;\; \theta=\;\frac{\pi}{2}

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