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  • For hyperbola    <img alt="\mathrm {\frac{x^2}{\cos ^2 \alpha}-\frac{y^2}{\sin ^2 \alpha}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%20%7B%5Cfrac%7Bx%5E2%7D%7

For hyperbola    \mathrm {\frac{x^2}{\cos ^2 \alpha}-\frac{y^2}{\sin ^2 \alpha}=1}   which of the following remains constant with change in  \mathrm { ' \alpha '} 
 

Option: 1

abscissae of vertices
 


Option: 2

abscissae of foci
 


Option: 3

eccentricity
 


Option: 4

directrix


Answers (1)

best_answer

The given equation of hyperbola is  \mathrm {\frac{x^2}{\cos ^2 \alpha}-\frac{y^2}{\sin ^2 \alpha}=1} 


\begin{aligned} & \mathrm {\Rightarrow a=\cos \alpha, b=\sin \alpha }\\ & \mathrm {\Rightarrow e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\tan ^2 \alpha}=\sec \alpha} \\ & \Rightarrow \mathrm {a e}=1 \\ & \therefore \quad \text { Foci }( \pm 1,0) \\ & \end{aligned}
\therefore \quad  Foci remain constant with respect to  \alpha.

Posted by

Suraj Bhandari

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