Question

An ellipse has a focus at $(ae,0)$ and directrix along $x=\frac{a}{e}$. If $b^2=a^2(1-e^2)$, then the equation of ellipse is

• A. x2a2+y2b2=1
• B. x2a2-y2b2=1
• C. x2a2+y2b2=0
• D. x2a2+y2b2=-1

Answer 1

The correct option is A x2a2+y2b2=1

As per definition of ellipse, Let the point be $S(h,k),$ focus be $(p,q)$ and directrix be $lx+my+n=0$
$\Rightarrow \frac{\text{Distance between}S\&(p,q)}{\text{Distance from}lx+my+n=0}=e$
$\Rightarrow {(h-p)}^2+{(k-q)}^2=e^2{\left(\frac{hl+mk+nl}{\sqrt{l^2+m^2}}\right)}^2$
$\Rightarrow (l^2+m^2)[{(h-p)}^2+{(k-q)}^2]=e^2{(lh+mk+n)}^2$
Here $l=1,m=0,n=\frac{-a}{e},p=ae,q=0$, we get,
$[{(h-ae)}^2+{(k-0)}^2]=e^2{(h+(0)k+\left(\frac{-a}{e}\right))}^2$
$\Rightarrow h^2+k^2={\left(eh\right)}^2+a^2(1-e^2)$
$\Rightarrow \frac{h^2}{a^2}+\frac{k^2}{a^2}(1-e^2)=1$
Given, $b^2=a^2(1-e^2)$
$\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$