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If the line $2 x+\sqrt{6} y=2$ touches the hyperbola $x^2-2 y^3=4$, then the coordinates of the point of contact are
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Verified Answer
The correct answer is:
$(4,-\sqrt{6})$
$2 x+\sqrt{6} y=2$
$\Rightarrow x^2-2 y^2=4$
Let point of contact $=\left(x_1, y_1\right)$
Equation of tangent at $\left(x_1, y_1\right)$
$=x x_1-2 y y_1-4=0$
$\because$ It represents same line as $2 x+\sqrt{6} y-2=0$
$\begin{array}{ll}\therefore & \frac{x_1}{2}=\frac{-2 y_1}{\sqrt{6}}=\frac{-4}{-2} \\ \therefore & x_1=4, y_1=-\sqrt{6} .\end{array}$
$\Rightarrow x^2-2 y^2=4$
Let point of contact $=\left(x_1, y_1\right)$
Equation of tangent at $\left(x_1, y_1\right)$
$=x x_1-2 y y_1-4=0$
$\because$ It represents same line as $2 x+\sqrt{6} y-2=0$
$\begin{array}{ll}\therefore & \frac{x_1}{2}=\frac{-2 y_1}{\sqrt{6}}=\frac{-4}{-2} \\ \therefore & x_1=4, y_1=-\sqrt{6} .\end{array}$
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