Question

Find the eccentricity, coordinates of foci, length of the latus-rectum of the following ellipse:
(i)$4x^2+9y^2=1$
(ii)$5x^2+4y^2=1$
(iii)$4x^2+3y^2=1$
(iv)$25x^2+16y^2=1600$
(v)$9x^2+25y^2=225$

Answer 1

(i) $4x^2+9y^2=1$
$\Rightarrow \frac{x^2}{\frac{1}{4}}+\frac{y^2}{\frac{1}{9}}=1$
This is in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$
where $a^2=\frac{1}{4}$ and $b^2=\frac{1}{9}$, i.e., $a=\frac{1}{2}$ and $b=\frac{1}{3}$
Clearly a>b
Now, $e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow e=\sqrt{1-\frac{\frac{1}{9}}{\frac{1}{4}}}$
$\Rightarrow e=\sqrt{1-\frac{4}{9}}$
$\Rightarrow e=\frac{\sqrt{5}}{3}$
Coordinates of the foci = $(\pm ae,0)=(\pm \frac{\sqrt{5}}{6},0)$
Length of the latus rectum= $\frac{2b^2}{a}$
$=\frac{2\times \frac{1}{9}}{\frac{1}{2}}=\frac{4}{9}$
(ii) $5x^2+4y^2=1$
$\Rightarrow \frac{x^2}{\frac{1}{5}}+\frac{y^2}{\frac{1}{4}}=1$
This is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
where $a^2=\frac{1}{5}$ and $b^2=\frac{1}{4},$ i.e.,
$a=\frac{1}{\sqrt{5}}$ and $b=\frac{1}{2}$
Clearly b>a
Now, $e=\sqrt{1-\frac{a^2}{b^2}}$
$\Rightarrow e=\sqrt{1-\frac{\frac{1}{5}}{\frac{1}{4}}}$
$\Rightarrow e=\sqrt{1-\frac{4}{5}}$
$\Rightarrow e=\frac{1}{\sqrt{5}}$
Coordinates of the foci
$=(0,\pm be)=(0,\pm \frac{1}{2\sqrt{5}})$
Length of the latus rectum =$\frac{2a^2}{b}$
$=\frac{2\times \frac{1}{5}}{\frac{1}{2}}$
$=\frac{4}{5}$
(iii) We have,
$4x^2+3y^2=1$
$\Rightarrow \frac{x^2}{\frac{1}{4}}+\frac{y^2}{\frac{1}{3}}=1$ ...(i)
This is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ where $a^2=\frac{1}{4}$ and $b^2=\frac{1}{3}$ i.e.,
$a=\frac{1}{2}$ and $b=\frac{1}{\sqrt{3}}$
Clearly, b>a, therefore the major and minor axes of the ellipse (i) are along y and x axes respectively.
Let e be the eccentricity of the ellipse. Then,
$e=\sqrt{1-\frac{a^2}{b^2}}$
$=\sqrt{1-\frac{\frac{1}{4}}{\frac{1}{3}}}$
$=\sqrt{1-\frac{3}{4}}$
$=\sqrt{\frac{1}{4}}$
$\therefore e=\frac{1}{2}$
The coordinates of the foci are (0, be) and (0,-be) i.e., $(0,\frac{1}{2\sqrt{3}})$ and $(0,\frac{-1}{2\sqrt{3}})$
Now,
Length of the latus rectum = $\frac{2a^2}{b}$
$=2\times \frac{\frac{1}{4}}{\frac{1}{\sqrt{3}}}$
$=\frac{\sqrt{3}}{2}$
(iv) We have,
$25x^2+16y^2=1600$
$\Rightarrow \frac{25x^2}{1600}+\frac{16y^2}{1600}=1$ $\Rightarrow \frac{x^2}{64}+\frac{y^2}{100}=1$
This is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ where $a^2=64$ and $b^2=100$ i.e.,
a=8 and b=10
Clearly, b>a, therefore the major and minor axes of the ellipse (i) are along y and x axes respectively.
Let e be the eccentricity of the ellipse. Then,
$e=\sqrt{1-\frac{a^2}{b^2}}$
$=\sqrt{1-\frac{64}{100}}$
$=\sqrt{\frac{36}{100}}$
$=\frac{6}{10}$
$=\frac{3}{5}$
The coordinates of the foci are (0,be) and (0,-be) i.e., (0,6) and (0,-6).
Now,
Length of the latus rectum = $\frac{2a^2}{b}$
$=2\times \frac{64}{10}$
$=\frac{64}{5}$
(v) $9x^2+25y^2=225$
$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$
This is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where
$a^2=25$ and $b^2=9$, i.e., $a=5$ and $b=3$
Clearly, $a>b$
Now, $e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow e=\sqrt{1-\frac{9}{25}}$
$\Rightarrow e=\sqrt{\frac{16}{25}}$
$\Rightarrow e=\frac{4}{5}$
Coordinates of the foci $=(\pm ae,0)=(\pm 4,0)$
Length of the latus rectum = $\frac{2b^2}{a}$
$=\frac{2\times 9}{5}$
$=\frac{18}{5}$