JEE Main 2019 — Hyperbola Question with Solution

Given hyperbola,

$$\frac{x^2}{{\cos }^2\theta }-\frac{y^2}{{\sin }^2\theta }=1$$

here a = cos$$\theta$$

and b = sin$$\theta$$

We know, eccentricity of the hyperbola is,

$$\sqrt{1+\frac{b^2}{a^2}}$$

$$\therefore$$  Here eccentricity

(e) = $$\sqrt{1+\frac{{\sin }^2\theta }{{\cos }^2\theta }}$$

Given that,

$$\sqrt{1+\frac{{\sin }^2\theta }{{\cos }^2\theta }}>2$$

$$\Rightarrow$$  $$\sqrt{1+{\tan }^2\theta }>2$$

$$\Rightarrow$$   1 + tan²$$\theta$$ > 4

$$\Rightarrow$$  tan²$$\theta$$ > 3

$$\Rightarrow$$  tan$$\theta$$ > $$\pm \sqrt{3}$$

As given $$\theta$$ $$\in$$ $$\left(0,\frac{\pi }{2}\right)$$

possible value of tan$$\theta$$ > $$\sqrt{3}$$

So, $$\theta$$ can be in the range $$\frac{\pi }{3}<\theta <\frac{\pi }{2}$$



We know latus ractum (LR) = $$\frac{2b^2}{a}$$

$$\therefore$$  LR = $$\frac{2{\sin }^2\theta }{\cos \theta }$$

= 2 tan$$\theta$$ sin$$\theta$$

We know in the range $$\frac{\pi }{3}<\theta <\frac{\pi }{2}$$ tan$$\theta$$ and sin$$\theta$$ both are increasing function.

So, at $$\frac{\pi }{3}$$ value of LR will be minimum and at $$\frac{\pi }{2}$$ value of LR will be maximum.

$$\therefore$$  Minimum value of LR = 2tan$$\frac{\pi }{3}$$ sin$$\frac{\pi }{3}$$

= 2 $$\times$$ $$\sqrt{3}\times \frac{\sqrt{3}}{2}$$
= 3

Maximum value of LR = 2tan$$\frac{\pi }{2}$$ sin$$\frac{\pi }{2}$$

= 2$$\left(\infty \right)\times 1$$

= $$\infty$$

$$\therefore$$  Interval of LR = (3, $$\infty$$)