Hyperbola
Equation of Hyperbola
IIT-JEE 2007
Grade 11
Question:
A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3x^2 + 4y^2 = 12$. Then its equation is:
(1) $x^2 \csc^2 \theta - y^2 \sec^2 \theta = 1$
(2) $x^2 \sec^2 \theta - y^2 \csc^2 \theta = 1$
(3) $x^2 \sin^2 \theta - y^2 \cos^2 \theta = 1$
(4) $x^2 \cos^2 \theta - y^2 \sin^2 \theta = 1$
Step-by-Step Solution
Key Concept: Ellipse $x^2/4 + y^2/3 = 1$: $c^2 = 1$, foci $(\pm 1, 0)$. Hyperbola transverse axis $2a_H = 2 \sin \theta \Rightarrow a_H = \sin \theta$. Same foci: $c_H = 1 \Rightarrow b_H^2 = c_H^2 - a_H^2 = 1 - \sin^2 \theta = \cos^2 \theta$. Equation: $x^2/\sin^2 \theta - y^2/\cos^2 \theta = 1$, i.e., $x^2 \csc^2 \theta - y^2 \sec^2 \theta = 1$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)