Question

If the line $2x+\sqrt{6}y=2$ touches the hyperbola $x^2-2y^2=4$, then the point of contact is

• A. $(-2,\sqrt{6})$
• B. $(-5,2\sqrt{6})$
• C. $(\frac{1}{2},\frac{1}{\sqrt{6}})$
• D. $(4,-\sqrt{6})$

Answer 1

Answer: (D)

The equation of tangent at $\left(x_1,y_1\right)$ is $x{x}_1-2y{y}_1=4$,
which is same as $2x+\sqrt{6}y=2$.
$\therefore \frac{x_1}{2}=-\frac{2y_1}{\sqrt{6}}=\frac{4}{2}$
$\Rightarrow x_1=4$ and $y_1=-\sqrt{6}$
Thus, the point of contact is $(4,-\sqrt{6})$.