Q. If the line $2x + \sqrt{6}y = 2$ touches the hyperbola $x^{2} - 2y^{2} = 4$, then the point of contact is
A) $\left( - 2,\ \sqrt{6} \right)$
B) $( - 5,\ 2\sqrt{6)}$
C) $\left( \frac{1}{2},\frac{1}{\sqrt{6}} \right)$
D) $(4, - \sqrt{6)}$
Explanation
(d)
As we know equation of tangent to the given hyperbola at ${(x}_{1,}y_{1})$ is ${xx}_{1} - 2{yy}_{1} = 4$ which is same as $2x + \sqrt{6}y = 2$
$\Rightarrow x_{1} = 4$ and $y_{1} = \sqrt{6}$
Thus, the point of contact is $(4,\ - \sqrt{6})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.