Question

The line $2x+\sqrt{6y}=2$ is a tangent to the curve $x^{2}-2y^{2}=4$. The point of contact is

• A. $\left(4,-\sqrt{6}\right)$
• B. $\left(7,-2\sqrt{6}\right)$
• C. $\left(2,3\right)$
• D. $\left(\sqrt{6},1\right)$

Answer 1

Answer: (A)

$\left(4,-\sqrt{6}\right)$

Answer 2

On solving the equation of line and curve, we get,
x^{2}-2\left\\{\frac{2-2 x}{\sqrt{6}}\right\\}^{2}=4
$\Rightarrow x^{2}-\frac{1}{3} \times 4\left(1+x^{2}-2 x\right)=4$
$\Rightarrow 3 x^{2}-4-4 x^{2}+8 x=12$
$\Rightarrow -x^{2}+8 x-16=0$
$\Rightarrow x^{2}-8 x+16=0$
$\Rightarrow (x-4)^{2}=0 \Rightarrow x=4$
and $\sqrt{6} \cdot y=2-2(4)=-6$
$\Rightarrow y=-\sqrt{6}$
So, point of contact is $(4,-\sqrt{6})$.