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Let $0 < \theta < \frac{\pi}{2}$ . If the eccentricity of the hyperbola $\frac{x}{\cos^2\theta} - \frac{y}{\sin^2\theta} = 1$ is greater than 2, then the length of its latus rectum lies in the interval:
Option 1)
$(3, \infty)$
Option 2)
$\left(\frac{3}{2}, 2 \right ]$
Option 3)
$(2,3]$
Option 4)
$\left (1, \frac{3}{2} \right ]$

Answers (1)

best_answer

Hyperbola -

Hyperbola is locus of all the points in a plane ,the difference of whose distance from two fixed point is constant.

- wherein

Length of latus Rectum -

$\frac{2b^{2}}{a}$

- wherein

For the Hyperbola

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

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

$given,\; \; \; \; e>2$

Eccentricity of the hyperbola is

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

$\therefore e=\sqrt{1+\tan ^{2}\theta }=\sec \left ( \theta \right )$

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

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

Latus rectum

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

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

This is strictly increasing

Latus Rectum $\equiv \left ( 3,00 \right )$

Option 1)
$(3, \infty)$
Option 2)

$\left(\frac{3}{2}, 2 \right ]$

Option 3)

$(2,3]$

Option 4)

$\left (1, \frac{3}{2} \right ]$

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