Question

The line $2x+\sqrt{6}y=2$ touches the hyperbola $x^2-2y^2=4$ at :

• A. $(4,-\sqrt{6})$
• B. $(2,-2\sqrt{6})$
• C. $(-4,\sqrt{6})$
• D. $(-2,2\sqrt{6})$

Answer 1

The correct option is A $(4,-\sqrt{6})$

Given hyperbola is $x^2-{2y}^2=4\Rightarrow \frac{x^2}{4}-\frac{y^2}{2}=1$

$a^2=4,b^2=2$

The condition of tangency is $a^2{l}^2-b^2{m}^2=n^2$

So from the given line we get $l=2,m=\sqrt{6},n=-2$

Substiute values we get, $4\times 4-2\times 6=4\Rightarrow 4=4$

So tangency condition holds true.

So line touches the hyperbola.

Now point of contact is $(-a^2\frac{l}{n},b^2\frac{m}{n})$

So we get $(-4\frac{2}{-2},2\frac{\sqrt{6}}{-2})=(4,-\sqrt{6})$