Question

Find the eccentricity, coordinates of foci, length of the latus-rectum of the following ellipse:
(i) 4x² + 9y² = 1
(ii) 5x² + 4y² = 1
(iii) 4x² + 3y² = 1
(iv) 25x² + 16y² = 1600.

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)4x^2+9y^2=1 \\ \Rightarrow \frac{x^2}{{\displaystyle \frac{1}{4}}}+\frac{y^2}{{\displaystyle \frac{1}{9}}}=1 \\ \mathrm{This}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{where}{a}^2=\frac{1}{4}\mathrm{and}{b}^2=\frac{1}{9},\mathrm{i}.\mathrm{e}.a=\frac{1}{2}\mathrm{and}b=\frac{1}{3}. \\ \mathrm{Clearly}a>b \\ \mathrm{Now},e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow e=\sqrt{1-\frac{{\displaystyle \frac{1}{9}}}{{\displaystyle \frac{1}{4}}}} \\ \Rightarrow e=\sqrt{1-\frac{4}{9}} \\ \Rightarrow e=\frac{\sqrt{5}}{3} \\ \mathrm{Coordinates}\mathrm{of}\mathrm{the}\mathrm{foci}=\left(\pm ae,0\right)=\left(\pm \frac{\sqrt{5}}{6},o\right) \\ \text{Length of the latus rectum=}\frac{2b^2}{a} \\ =\frac{2\times {\displaystyle \frac{1}{9}}}{{\displaystyle \frac{1}{2}}} \\ =\frac{4}{9} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right)5x^2+4y^2=1 \\ \Rightarrow \frac{x^2}{{\displaystyle \frac{1}{5}}}+\frac{y^2}{{\displaystyle \frac{1}{4}}}=1 \\ \mathrm{This}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{where}{a}^2=\frac{1}{5}\mathrm{and}{b}^2=\frac{1}{4},\mathrm{i}.\mathrm{e}.a=\frac{1}{\sqrt{5}}\mathrm{and}b=\frac{1}{2}. \\ \mathrm{Clearly}b>a \\ \mathrm{Now},e=\sqrt{1-\frac{a^2}{b^2}} \\ \Rightarrow e=\sqrt{1-\frac{{\displaystyle \frac{1}{5}}}{{\displaystyle \frac{1}{4}}}} \\ \Rightarrow e=\sqrt{1-\frac{4}{5}} \\ \Rightarrow e=\frac{1}{\sqrt{5}} \\ \mathrm{Coordinates}\mathrm{of}\mathrm{the}\mathrm{foci}=\left(0,\pm be\right)=\left(0,\pm \frac{1}{2\sqrt{5}}\right) \\ \text{Length of the latus rectum=}\frac{2a^2}{b} \\ =\frac{2\times {\displaystyle \frac{1}{5}}}{{\displaystyle \frac{1}{2}}} \\ =\frac{4}{5} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)4x^2+3y^2=1 \\ \Rightarrow \frac{x^2}{{\displaystyle \frac{1}{4}}}+\frac{y^2}{{\displaystyle \frac{1}{3}}}=1 \\ \mathrm{This}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{where}{a}^2=\frac{1}{4}\mathrm{and}{b}^2=\frac{1}{3},\mathrm{i}.\mathrm{e}.a=\frac{1}{2}\mathrm{and}b=\frac{1}{\sqrt{3}}. \\ \mathrm{Clearly},b>a \\ \mathrm{Now},e=\sqrt{1-\frac{a^2}{b^2}} \\ \Rightarrow e=\sqrt{1-\frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{3}}}} \\ \Rightarrow e=\sqrt{1-\frac{3}{4}} \\ \Rightarrow e=\frac{1}{2} \\ \mathrm{Coordinates}\mathrm{of}\mathrm{the}\mathrm{foci}=\left(0,\pm be\right)=\left(0,\pm \frac{1}{2\sqrt{3}}\right) \\ \text{Length of the latus rectum=}\frac{2a^2}{b} \\ =\frac{2\times {\displaystyle \frac{1}{4}}}{{\displaystyle \frac{1}{\sqrt{3}}}} \\ =\frac{\sqrt{3}}{2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)25x^2+16y^2=1600 \\ \Rightarrow \frac{x^2}{{\displaystyle 64}}+\frac{y^2}{{\displaystyle 100}}=1 \\ \mathrm{This}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\mathrm{where}{a}^2=64\mathrm{and}{b}^2=100,\mathrm{i}.\mathrm{e}.a=8\mathrm{and}b=10. \\ \mathrm{Clearly},b>a \\ \mathrm{Now},e=\sqrt{1-\frac{a^2}{b^2}} \\ \Rightarrow e=\sqrt{1-\frac{{\displaystyle 64}}{{\displaystyle 100}}} \\ \Rightarrow e=\sqrt{\frac{36}{100}} \\ \Rightarrow e=\frac{6}{10}\mathrm{or}\frac{3}{5} \\ \mathrm{Coordinates}\mathrm{of}\mathrm{the}\mathrm{foci}=\left(0,\pm 6\right) \\ \text{Length of latus rectum=}\frac{2a^2}{b} \\ =\frac{2\times {\displaystyle 64}}{{\displaystyle 10}} \\ =\frac{64}{5} \end{gathered}$$