Question

Find the equation of the ellipse having the latus recta of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as tangents and the point $\left(0,\pm b\right)$ as its focii.

Answer 1

Where $e=\sqrt{1-\frac{b^2}{a^2}}$ for ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

For new ellipse,

Let eccentricity be e'

$\therefore a^{\prime }{e}^{\prime }=b$

and $e^{\prime }=\sqrt{1-\frac{a^2{e}^2}{a^{\prime 2}}}$

$\Rightarrow a^{\prime }{e}^{\prime }=\sqrt{a^{\prime 2}-a^2{e}^2}$ $\therefore \frac{x^2}{a^2-b^2}+\frac{y^2}{a^2}=1$

$\Rightarrow b^2+a^2{e}^2=a^{\prime 2}$

$\Rightarrow a^{\prime 2}=b^2+(a^2-b^2)=a^2$.