Question

If foci of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ concide with the foci of $\frac{x^2}{25}+\frac{y^2}{9}=1$ and eccentricity of the hyperbola is $2$, then

• A. $a^2+b^2=16$
• B. end points of latus rectum will be $(\pm 4,\pm 6)$
• C. ratio of lengths of transverse axis to conjugate axis is $1:\sqrt{3}$
• D. length of latus rectum of the hyperbola $=12$ unit

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $a^2+b^2=16$
B end points of latus rectum will be $(\pm 4,\pm 6)$
C ratio of lengths of transverse axis to conjugate axis is $1:\sqrt{3}$
D length of latus rectum of the hyperbola $=12$ unit

For the ellipse :$a=5,e=\sqrt{1-\frac{9}{25}}=\frac{4}{5}$
$\therefore ae=4$
$\therefore$ The foci are $(-4,0)$ and $(4,0)$
For the hyperbola
$ae=4,e=2$
$\therefore a=2$
$b^2=4(4-1)=12$
Hence, the ratio of transverse axis to conjugate axis is $1:\sqrt{3}$
Clearly, $a^2+b^2=4+12=16$
end points of latus rectum are $(\pm ae,\pm \frac{b^2}{a})=(\pm 4,\pm 6)$
Length of latus rectum
$=\frac{2b^2}{a}=\frac{2\times 12}{2}=12$ unit