Hyperbola 2 Question 10
10. If $2 x-y+1=0$ is a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{16}=1$ then which of the following CANNOT be sides of a right angled triangle?
(2017 Adv.)
(a) $a, 4,1$
(b) $2 a, 4,1$
(c) $a, 4,2$
(d) $2 a, 8,1$
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Answer:
Correct Answer: 10. $(a, b)$
Solution:
- Tangent $\equiv 2 x-y+1=0$
Hyperbola $\equiv \frac{x^{2}}{a^{2}}-\frac{y^{2}}{16}=1$
It point $\equiv(a \sec \theta, 4 \tan \theta)$
$$ \text { tangent } \equiv \frac{x \sec \theta}{a}-\frac{y \tan \theta}{4}=1 $$
On comparing, we get $\sec \theta=-2 a$
$$ \begin{aligned} & & \tan \theta=-4 \Rightarrow 4 a^{2}-16=1 \\ \therefore & a & =\frac{\sqrt{17}}{2} \end{aligned} $$
Substitute the value of $a$ in option (a), (b), (c) and (d).
