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VITEEE 2009 Paper

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Question : 106 of 120

Marks: +1, -0

The mid point of the chord 4x - 3y = 5 of the hyperbola

2x^{2}-3y^{2}=12

is

$\left(0,-\frac{5}{3}\right)$
(2,1)
$\left(\frac{5}{4},0\right)$
$\left(\frac{11}{4},2\right)$

Solution:

Solving

4x-3y=5

and

2x^{2}-3y^{2}=12

∴

2\left(\frac{5+3y}{4}\right)^{2}-3y^{2}=12

⇒

\frac{25+9y^{2}+30y}{8}-3y^{2}=12\Rightarrow 15y^{2}-30y+71=0

⇒

y=\frac{30\pm\sqrt{900-4260}}{30}=1\pm\frac{\sqrt{-3360}}{30}

Also,

2x^{3}-3\left(\frac{4x-5}{3}\right)^{2}=12\Rightarrow 10x^{2}-40x+61=0

⇒

x=\frac{40\pm\sqrt{1600-4\times10\times61}}{2\times10}

=\frac{40\pm\sqrt{-840}}{20}=2\pm\frac{\sqrt{-840}}{20}

∴ Points are

A\left(2+\frac{\sqrt{-840}}{20},1+\frac{\sqrt{-3360}}{30}\right)

and

B\left(2-\frac{\sqrt{-840}}{20},1-\frac{\sqrt{-3360}}{30}\right)

.

∴ Mid point of AB is (2,1).

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