Question

Find the equation of an ellipse whose eccentricity is 2/3, the latus-rectum is 5 and the centre is at the origin.

Answer 1

$$\begin{gathered} \text{Let the ellipse be}\frac{x^2}{a^2}\text{+}\frac{y^2}{b^2}\text{=1. ...}\left(\text{1}\right) \\ e=\frac{2}{3}\mathrm{and}\mathrm{l}\text{atus rectum=5 (Given)} \\ \mathrm{Now},\frac{2b^2}{a}=5 \\ \Rightarrow 2b^2=5a...\left(2\right) \\ \Rightarrow 2a^2(1-e^2)=5a[\because b^2=a^2(1-e^2\left)\right] \\ \Rightarrow 2a^2\left[1-\frac{4}{9}\right]=5a \\ \Rightarrow 2a^2\times \frac{5}{9}=5a \\ \Rightarrow 10a^2=45a \\ \Rightarrow a=\frac{9}{2} \\ \text{Substituting the value of}\mathit{\text{a}}\text{in eq. (2), we get:} \\ 2b^2=5\times \frac{9}{2} \\ \Rightarrow b^2=\frac{45}{4} \\ \text{Substituting the values of}{\mathit{\text{a}}}^2\text{and}{\mathit{\text{b}}}^{\mathit{2}}\text{in eq. (1), we get:} \\ \frac{x^2}{{\displaystyle \frac{81}{4}}}\text{+}\frac{y^2}{{\displaystyle \frac{45}{4}}}\text{=1} \\ \Rightarrow \frac{4x^2}{81}+\frac{4y^2}{45}=1 \\ \mathrm{This}\text{is the required equation of the ellipse.} \end{gathered}$$