For the hyperbola $\frac{x^{2}}{\cos ^{2} \alpha}-\frac{y^{2}}{\sin ^{2} \alpha}=1$, which of the following remains constant when $\alpha$ varies
- Abscissae of vertices
- Abscissae of foci
- Eccentricity
- Directrix
The Correct Option is B
Solution and Explanation
Given Hyperbola \(\frac{x^{2}}{\cos ^{2} \alpha}-\frac{y^{2}}{\sin ^{2} \alpha}=1\)
given that the angle \(\alpha\) varies
We have \(a=\cos \alpha, b=\sin \alpha\)
now eccentricity \(e=\frac{\sqrt{a^{2}+b^{2}}}{a}=\frac{\sqrt{\cos ^{2} \alpha+\sin ^{2} \alpha}}{\cos \alpha}=\frac{1}{\cos \alpha}\)
Thus eccentricity varies with \(\alpha\).
Now foci \((\pm a e, 0)=(1,0)\) Independent of \(\alpha\).
Vertex \((\pm a, 0)=(\cos \alpha, 0)\) dependent on \(\alpha\).
Directrix is given by \(x=\pm \frac{a}{e}=\cos ^{2} \alpha\)
so dependent on \(\alpha\).
So abscissae of foci are independent of \(\alpha\).
The correct answer is (B): Abscissae of foci
Top Questions on Hyperbola
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Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
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