Question

Find the latus rectum, the eccentricity, and the coordinates of the foci, of the ellipses
(1) $x^2+3y^2=a^2$, (2) $5x^2+4y^2=1$, and (3) $9x^2+5y^2-30y=0$.

Answer 1

(1) $x^2+3y^2=a^2$

$\frac{x^2}{a^2}+\frac{y^2}{\left(\frac{a^2}{3}\right)}=1$

Latus rectum$=\frac{2b^2}{a}=\frac{2\times \frac{a^2}{3}}{a}=\frac{2a}{3}$

$e^2=1-\frac{b^2}{a^2}=1-\frac{\frac{a^2}{3}}{a^2}=1-\frac{1}{3}=\frac{2}{3}\Rightarrow e=\sqrt{\frac{2}{3}}=\frac{1}{3}\sqrt{6}$

Coordinates of foci are $(\pm ae,0)$

$(\pm \frac{a}{3}\sqrt{6},0)$

(2) $5x^2+4y^2=1$

$\frac{x^2}{{\left(\frac{1}{\sqrt{5}}\right)}^2}+\frac{y^2}{{\left(\frac{1}{2}\right)}^2}=1a=\frac{1}{\sqrt{5}},b=\frac{1}{2}b>a$

So the major axis of ellipse is $y$ axis

Latus rectum $=\frac{2a^2}{b}=\frac{2\times \frac{1}{5}}{\frac{1}{2}}=\frac{4}{5}$

$e^2=1-\frac{a^2}{b^2}=1-\frac{\frac{1}{5}}{\frac{1}{4}}=1-\frac{4}{5}=\frac{1}{5}\Rightarrow e=\frac{1}{\sqrt{5}}$

Foci : $(0,\pm be)$

$\Rightarrow (0,\pm \frac{1}{2\sqrt{5}})$

(3) $9x^2+5y^2-30y=0$

$9x^2+5(y^2-6y+9-9)=09x^2+5{(y-3)}^2=45\frac{x^2}{5}+\frac{{(y-3)}^2}{9}=1a=\sqrt{5},b=3b>a$

So, the major axis of ellipse is $y$ axis.

Latus rectum $=\frac{2a^2}{b}=\frac{2\times 5}{3}=\frac{10}{3}$

$e^2=1-\frac{a^2}{b^2}=1-\frac{5}{9}=\frac{4}{9}\Rightarrow e=\frac{2}{3}$

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

Foci is $(h,\pm be+k)$

Here $h=0$ and $k=3$

So foci are $(0,3\times \frac{2}{3}+3)=(0,5)$ and $(0,-3\times \frac{2}{3}+3)=(0,1)$