Question

Find the equation to the ellipse whose one vertex is $(3,1)$, the nearer focus is $(1,1)$ and the eccentricity is $2/3$.

Answer 1

$SV=\sqrt{4.10}$

$SV=2$

Equation of ellipse with centre $(h,k)$

$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$

Given

$SV=2$

$=a-ae=2$

$=a(1-\frac{2}{3})=2$

$\therefore a=6$

$e=\frac{2}{3}$

$\therefore \sqrt{1-\frac{b^2}{a^2}}=\frac{2}{3}\frac{b^2}{a^2}=\frac{5}{9}{b}^2=\frac{b{a}^2}{9}$

clearly $x=1$ axis of ellipse

$cs=ae$

$=1-\lambda =6\times \frac{2}{3}$

$=\lambda =-3$

$\therefore c(-3,1)$

Equation of ellipse with $C(-3,1)$ is

$\frac{(x+3{)}^2}{36}+\frac{(y-1{)}^2}{\frac{5}{9}{a}^2}=1$

$=\frac{(x+3{)}^2}{36}+\frac{(y-1{)}^2}{20}=1$