Question

Find the length of axes of the ellipse whose eccentricity is 4/5 and whose foci eoincide with those of the hyperbola $9x^2-16y^2+144=0$

Answer 1

$9x^2-16y^2+144=0$

$\frac{x^2}{4^2}-\frac{y^2}{3^2}=0$

Therefore, $a=4,b=3$

$3^2=4^2(e^2-1)$

$e=\frac{5}{4}$

Therefore, focii are $(5,0)$ and $(-5,0)$

For ellipse,

Length of major axis = $2a$

$=2\left(\frac{25}{4}\right)=\frac{25}{2}$ $[ae=5]$

Length of minor axis = $\frac{2b^2}{a}$

$=\frac{18}{4}=\frac{9}{2}$ $[b^2=a^2(1-e^2\left)\right]$