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

Answer 1

The correct option is A

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

Solve both equations to get the point of contact.

The given equation of line is $2x+\sqrt{6}y=2$.

or, $y=\frac{2-2x}{\sqrt{6}}$………..$\left(i\right)$

Equation of curve is $x^2-2y^2=4$.

Substitute $y$ in above equation

$\begin{array}{rcl}{x}^2-2\frac{{\left(2-2x\right)}^2}{6}& =& 4\\ 6x^2-2\left(4+4x^2-8x\right)& =& 24\\ 6x^2-8-8x^2+16x-24& =& 0\\ -2x^2+16x-32& =& 0\\ x^2-8x+16& =& 0\\ {\left(x-4\right)}^2& =& 0\end{array}$

So $x=4$

Put $x=4$ in $\left(i\right)$

$\begin{array}{rcl}y& =& \frac{2-8}{\sqrt{6}}\\ & =& \frac{-6}{\sqrt{6}}\\ & =& -\sqrt{6}\end{array}$

So, the point of contact is $\left(4,-\sqrt{6}\right)$.

Hence, option $\left(A\right)$ is correct.