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  4. If the line 2x+√6y=2 touches the hyperbola x2-2y2=4, then the point of contact is

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

IIT JEEIIT JEE 2004Conic Sections

Solution:

The equation of tangent at $\left(x_{1}, y_{1}\right)$ is $x x_{1}-2 y y_{1}=4$,
which is same as $2 x+\sqrt{6} y=2$.
$\therefore \frac{x_{1}}{2}=-\frac{2 y_{1}}{\sqrt{6}}=\frac{4}{2}$
$\Rightarrow x_{1}=4$ and $y_{1}=-\sqrt{6}$
Thus, the point of contact is $(4,-\sqrt{6})$.