COMEDK · Maths · 14. Ellipse
The foci of a hyperbola are the same as those of the ellipse with equation \(9x^2 + 16y^2 = 144\). If the length of the transverse axis of this hyperbola is \(2\cos\alpha\), then its equation is:
- A \(\dfrac{x^2}{\cos^2\alpha} - \dfrac{y^2}{5 - \cos^2\alpha} = 1\)
- B \(\dfrac{x^2}{7 - \cos^2\alpha} - \dfrac{y^2}{\cos^2\alpha} = 1\)
- C \(\dfrac{x^2}{\cos^2\alpha} - \dfrac{y^2}{7 + \cos^2\alpha} = 1\)
- D \(\dfrac{x^2}{\cos^2\alpha} - \dfrac{y^2}{7 - \cos^2\alpha} = 1\)
Answer & Solution
Correct Answer
(D) \(\dfrac{x^2}{\cos^2\alpha} - \dfrac{y^2}{7 - \cos^2\alpha} = 1\)
Step-by-step Solution
Detailed explanation
The equation of the given ellipse is \(9x^2 + 16y^2 = 144\), which can be written as \(\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1\). Here, \(a^2 = 16\) and \(b^2 = 9\). The eccentricity \(e\) of the ellipse is given by…
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