MHT CET · Maths · Ellipse
The foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\). The equation of the hyperbola with eccentricity 2 is
- A \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
- B \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
- C \(\frac{x^2}{12}-\frac{y^2}{16}=1\)
- D \(\frac{x^2}{16}-\frac{y^2}{12}=1\)
Answer & Solution
Correct Answer
(B) \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
Step-by-step Solution
Detailed explanation
For ellipse: \(a_e^2=25, b_e^2=9\) \(c_e^2 = a_e^2 - b_e^2 = 25 - 9 = 16 \Rightarrow c_e = 4\)
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