The foci of the hyperbola coincide with the ellipse x2/25 + y2/9 = 1. If the eccentricity of the hyperbola is 2. Find its equation

Question

The answer obtained is the hyperbola x2/4-y2/12 = 1. With b > a the foci do not coincide.... What is the correct answer

 

With b2 > a2, the hyperbola will be vertical. How will the foci coincide?

Answer 1

If we take the question to mean that the foci of the ellipse are also the foci of an hyperbola, then we have, for the ellipse, 

a² -c² = b² so 25 -c^{2 =} 9 and that means c² = 16.  This ellipse has its major axis on the x-axis.

 

For the hyperbola, which must have its transverse axis on the x-axis, the equation c² - a² = b² and e = c/a = 2. Only the c value is the same as for the ellipse; c = 4.  Thus 4/a=2 tells us that a (for the hyperbola) = 2.

Therefore we compute b² =16 - 4 = 12.

 

The equation of the hyperbola is x²/a² - y²/b² = 1; substituting gives us x² - y²/12 = 1.

THIS IS THE CORRECT ANSWER.