Question:
Which one of the following is independent of \(\alpha\) in the hyperbola \((0<\alpha<\pi/2)\) \(\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1\)?
Which one of the following is independent of \(\alpha\) in the hyperbola \((0<\alpha<\pi/2)\) \(\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1\)?
Show Hint
Check which parameter remains constant after substitution.
Updated On: Apr 23, 2026
- Eccentricity
- Abscissa of foci
- Directrix
- Vertex
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Formula / Definition}
\[ a^2 = \cos^2 \alpha, b^2 = \sin^2 \alpha, e^2 = 1 + \frac{b^2}{a^2} = 1 + \tan^2 \alpha = \sec^2 \alpha \]
Step 2: Calculation / Simplification}
\(a = \cos \alpha, e = \sec \alpha\)
Foci: \((\pm ae, 0) = (\pm \cos \alpha \sec \alpha, 0) = (\pm 1, 0)\)
Abscissa of foci = \(\pm 1\), independent of \(\alpha\).
Step 3: Final Answer
\[ \text{Abscissa of foci} \]
\[ a^2 = \cos^2 \alpha, b^2 = \sin^2 \alpha, e^2 = 1 + \frac{b^2}{a^2} = 1 + \tan^2 \alpha = \sec^2 \alpha \]
Step 2: Calculation / Simplification}
\(a = \cos \alpha, e = \sec \alpha\)
Foci: \((\pm ae, 0) = (\pm \cos \alpha \sec \alpha, 0) = (\pm 1, 0)\)
Abscissa of foci = \(\pm 1\), independent of \(\alpha\).
Step 3: Final Answer
\[ \text{Abscissa of foci} \]
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