Question

The equation of tangents to the hyperbola $3x^2-2y^2=6$, which is perpendicular to the line $x-3y=3$, are

• A. $y=-3x\pm \sqrt{15}$
• B. $y=3x\pm \sqrt{6}$
• C. $y=-3x\pm \sqrt{6}$
• D. $y=2x\pm \sqrt{15}$

Answer 1

Answer: (A)

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

Equation of hyperbola is
$3x^2-2y^2=6$
$\Rightarrow \frac{x^2}{2}-\frac{y^2}{3}=1$
So, $a^2=2$ and $b^2=3$
Given, equation of line is $x-3y=3$

$⟹y=\frac{1}{3}x+1$
$\therefore$ Slope of given line $=\frac{1}{3}$
$\therefore$ Slope of line perpendicular to given line is
$m=-3$
The equation of tangents is given by
$y=mx\pm \sqrt{a^2{m}^2-b^2}$
$=-3x\pm \sqrt{2\times 9-3}$
$=-3x\pm \sqrt{18-3}$
$=-3x\pm \sqrt{15}$

Hence, option A is correct.