Question

If the line $2x+\sqrt6y=2$ touches the hyperbola $x^2-2y^2=4$, then the point of contact is

• A. $(-2,\sqrt6)$
• B. $(-5,2\sqrt6)$
• C. $\Bigg(\frac{1}{2},\frac{1}{\sqrt6}\Bigg)$
• D. $(4,-\sqrt6)$

Answer 1

Answer: (D)

$(4,-\sqrt6)$

Answer 2

The equation of tangent at $(x_1,y_1)$ is $xx_1-2yy_1=4$,which is same as $2x+\sqrt6y=2.$
$\therefore\, \, \, \, \, \, \, \, \frac{x_1}{2}=-\frac{2y_1}{\sqrt6}=\frac{4}{2}$
$\Rightarrow\, \, \, \, \, \, \, \, \, x_1=4$ and $y_1=-\sqrt6$
Thus, the point of contact is $(4,-\sqrt6)$.