Question

The foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. Then the equation 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. $3x^2-y^2+12=0$
• D. $9x^2-25y^2-225=0$

Answer 1

The correct option is B $\frac{x^2}{4}-\frac{y^2}{12}=1$

The foci of the ellipse are also the foci of an hyperbola,

then we have, for the ellipse,

$a^2-c^2=b^2$ so $25-c^2=9$ and that means $c^2=16.$

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

$c^2-a^2=b^2$ and

$e=\frac{c}{a}=2.$

Only the $c$ value is the same as for the ellipse; $c=4.$

Thus $\frac{4}{a}=2$

tells us that $a$ (for the hyperbola) $=2.$

Therefore we compute $b^2=16-4=12.$

The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ substituting gives us $\frac{x^2}{4}-\frac{y^2}{12}=1$