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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

  1. A \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
  2. B \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
  3. C \(\frac{x^2}{12}-\frac{y^2}{16}=1\)
  4. D \(\frac{x^2}{16}-\frac{y^2}{12}=1\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{x^2}{4}-\frac{y^2}{12}=1\)

Step-by-step Solution

Detailed explanation

For ellipse: \(a_e^2=25, b_e^2=9\) \(c_e^2 = a_e^2 - b_e^2 = 25 - 9 = 16 \Rightarrow c_e = 4\)