Question

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

(IIT JEE 2004)

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

Answer 1

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

We have equation of hyperbola
$x^2-2y^2=4$
Equation of tangent to above hyperbola at point $(x_1,y_1)$ is
$x{x}_1-2y{y}_1=4$ ...(i)
Equation of tangent
Given,
$2x+\sqrt{6}y=2$
$\Rightarrow 4x+2\sqrt{6}y=4$ ...(ii)
Comparting eq(i) and (ii) we get
$x_1=4$
And, $2y_1=-2\sqrt{6}\Rightarrow y_1=-\sqrt{6}$
so, the point of contact is $(4,-\sqrt{6}).$