Question:
If the line $2x+\sqrt6y=2$ touches the hyperbola $x^2-2y^2=4$, then the point of contact is
If the line $2x+\sqrt6y=2$ touches the hyperbola $x^2-2y^2=4$, then the point of contact is
Updated On: Jun 14, 2022
- $(-2,\sqrt6)$
- $(-5,2\sqrt6)$
- $\Bigg(\frac{1}{2},\frac{1}{\sqrt6}\Bigg)$
- $(4,-\sqrt6)$
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The Correct Option is D
Solution and Explanation
The equation of tangent at $(x_1,y_1)$ is $xx_1-2yy_1=4$,which is same as $2x+\sqrt6y=2.$
$\therefore\, \, \, \, \, \, \, \, \frac{x_1}{2}=-\frac{2y_1}{\sqrt6}=\frac{4}{2}$
$\Rightarrow\, \, \, \, \, \, \, \, \, x_1=4$ and $y_1=-\sqrt6$
Thus, the point of contact is $(4,-\sqrt6)$.
$\therefore\, \, \, \, \, \, \, \, \frac{x_1}{2}=-\frac{2y_1}{\sqrt6}=\frac{4}{2}$
$\Rightarrow\, \, \, \, \, \, \, \, \, x_1=4$ and $y_1=-\sqrt6$
Thus, the point of contact is $(4,-\sqrt6)$.
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