Mathematics | VITEEE 2015 | Discussion Page

The equation of tangent to the hyperbola 3x²-2y²=6 , which is perpendicular to the line x-3y=3, are

$y=-3x\pm\sqrt{15}$

$y=3x\pm\sqrt{6}$

$y=-3x\pm\sqrt{6}$

$y=2x\pm\sqrt{15}$

Option A

Equation of hyperbola is

3x²-2y²=6

$\Rightarrow$ $\frac{x^{2}}{2}-\frac{y^{2}}{3}=1$

So, a² =2 and b²=3

Given, equation of line is x-3y=3

$\therefore$ Slope of given line = $\frac{1}{3}$

$\therefore$ Slope of line perpendicular to given line, m=-3

The equation of tangents are

$y=mx\pm\sqrt{a^{2}m^{2}-b^{2}}$

$=-3x\pm\sqrt{2\times9-3}$

$=-3x\pm\sqrt{18-3}$

$=-3x\pm\sqrt{15}$

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