Question

Let 0 < θ < π/2 . If the eccentricity of the hyperbola (x²/cos²θ) - (y²/sin²θ) = 1 is greater than 2, then the length of its latus rectum lies in the interval: (A) (3, [infinity]) (B) (3/2, 2]
(C) (2, 3] (D) (1, 3/2]
[JEE Main 2019]

Answer 1

The length of its Latus Rectum lies in the interval:

(A) (3, [infinity])

• We know that, the length of Latus Rectum is = $\frac{2b^{2}}{a}$

• Now, In case of hyperbola,

$\frac{x^{2}}{a^{2}}- \frac {y^{2}}{b^{2}}= 1$

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

• As per the question, e >2

• Thus, Eccentricity of the hyperbola is-

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

• Therefore, $e=\sqrt{1+\tan ^{2}\theta }$  

                             = sec Θ

• Also,    $\sec \left ( \theta \right )>2\Rightarrow \cos \left ( \theta \right )<\frac{1}{2}$

• Thus, $\theta \equiv \left ( 60^{\circ},90^{\circ} \right )$

• So, Latus Rectum =  LR =  $\frac{2b^{2}}{a}=\frac{2\sin ^{2}\theta }{\cos \theta }=\frac{2\left ( 1-\cos ^{2}\theta \right )}{\cos \theta }$

                                               $=2\sec \theta -2\cos \theta$

• Since, it is increasing, thus the range of Latus Rectum is (3, [infinity])