Find the eccentricity of an ellipse, if the length of its latus rectum is 4 units and distance between its…
Find the eccentricity of an ellipse, if the length of its latus rectum is 4 units and distance between its vertex and the nearest focus is \(3 / 2\) units.
\(\frac{1}{3}\)
\(\frac{2}{3}\)
\(\frac{1}{9}\)
\(\frac{3}{4}\)
Solution
Let the equation of an ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a > b)\) Then according to the question,
Length of latus rectum \(=\frac{2 b^2}{a}=4\) ...(i)
and \(\quad a-a e=\frac{3}{2} \Rightarrow a(1-e)=\frac{3}{2}\) ...(ii)
From Eqs. (i) and (ii), we get
\(\begin{aligned}
\frac{\frac{2 b^2}{a}}{a(1-e)} & =\frac{4}{3} \Rightarrow \frac{2 b^2}{a^2}=\frac{8}{3}(1-e) \\
\left(1-e^2\right) & =\frac{4}{3}(1-e) \quad\left\{\because \frac{b^2}{a^2}=1-e^2\right\} \\
\Rightarrow \quad 1+e & =\frac{4}{3} \Rightarrow e=\frac{4}{3}-1 \Rightarrow e=\frac{1}{3}
\end{aligned}\)
Hence, option (a) is correct.