Question

Consider an ellipse whose focus is at $(ae,0)$ and directrix along $(x=\frac{a}{e})$.
Find the equation of the ellipse if
$b^2=a^2(1-e^2)$

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
• B. $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
• C. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=0$
• D. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=-1$

Answer 1

The correct option is A

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

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