For the Hyperbola {x^2}{sec ^2}theta - {y^2}c | vedclass

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Question

Given equation is comparing on $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

We get

$a^{2}=\cos ^{2} a$ and $b^{2}=\sin ^{2} a$

$	herefore \si

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Given equation is comparing on $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

We get

$a^{2}=\cos ^{2} a$ and $b^{2}=\sin ^{2} a$

$	herefore \sin ^{2} a+\cos ^{2} a=a^{2}+b^{2}$

$\Rightarrow e-\frac{\overline{a^{2}+b^{2}}}{a^{2}}$

Now,

$=\frac{\overline{1}}{\cos ^{2} a}=\frac{1}{\cos a}$

Now, foci $a e=\cos a \cdot \frac{1}{\cos \alpha}=1$

For the Hyperbola ${x^2}{\sec ^2}\theta - {y^2}cose{c^2}\theta = 1$ which of the following remains constant when $\theta $ varies $= ?$

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