Question

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

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

Answer 1

Answer: (A)

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

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