Hyperbola 2 Question 8
8. If the line $2 x+\sqrt{6} y=2$ touches the hyperbola $x^{2}-2 y^{2}=4$, then the point of contact is
(2004, 1M)
(a) $(-2, \sqrt{6)}$
(b) $(-5,2 \sqrt{6})$
(c) $\frac{1}{2}, \frac{1}{\sqrt{6}}$
(d) $(4,-\sqrt{6})$
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Answer:
Correct Answer: 8. (a, c, d)
Solution:
- The equation of tangent at $\left(x _1, y _1\right)$ is $x x _1-2 y y _1=4$, which is same as $2 x+\sqrt{6} y=2$.
$$ \begin{array}{ll} \therefore & \frac{x _1}{2}=-\frac{2 y _1}{\sqrt{6}}=\frac{4}{2} \\ \Rightarrow & x _1=4 \text { and } y _1=-\sqrt{6} \end{array} $$
Thus, the point of contact is $(4,-\sqrt{6})$.
