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VITEEE 2014 Paper

Section: Mathematics

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Question : 111 of 120

Marks: +1, -0

The angle between lines joining the origin to the point of intersection of the line

\sqrt{3}

x + y = 2 and the curve

y^2-x^2

= 4 is

$\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$
$\frac{\pi}{6}$
$\tan^{-1}\left(\frac{\sqrt{3}}{\sqrt{2}}\right)$
$\frac{\pi}{2}$

Solution:

On homogenising

y^2 - x^2

= 4 with the

\sqrt{3}

x + y = 2 , we get

y^2-x^2

= 4

\frac{(\sqrt{3}x+y)^2}{4}

⇒

y^2-x^2

=

3x^2+y^2+2\sqrt{3}

xy

⇒

4x^2 + 2\sqrt{3}

xy =

On comparing with

ax^2

+ 2hxy +

by^2

= 0 ,

we get

a = 4 , h =

\sqrt{3}

and b = 0

We know that.

tan θ = 2

\frac{\sqrt{h^2-ab}}{a+b}

=

\frac{2\sqrt{3-0}}{4+0}

∴ The angle between the lines is

θ =

\tan^{-1}\left(\frac{\sqrt{3}}{\sqrt{2}}\right)

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