Question
Let an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{{\text{ }b}^{2}} = 1,a < b$, pass through the point $(4,3)$ and have eccentricity $\frac{\sqrt{5}}{3}$. Then the length of its latus rectum is :
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Solution
Sol. [Key Concept: For $a < b$ use $e^{2} = 1 - a^{2}/b^{2}$ with the given point to get $a^{2},b^{2}$; latus rectum $= 2a^{2}/b$.]
$$\ e^{2} = 1 - \frac{a^{2}}{{\text{ }b}^{2}} \Rightarrow \frac{a^{2}}{{\text{ }b}^{2}} = 1 - e^{2} = 1 - \frac{5}{9} = \frac{4}{9}$$
$$\begin{array}{r} \Rightarrow \frac{a^{2}}{{\text{ }b}^{2}} = \frac{4}{9}\#(1) \end{array}$$
Passes through $(4,3) \Rightarrow \frac{16}{a^{2}} + \frac{9}{{\text{ }b}^{2}} = 1$
from (1) and (2)
$a^{2} = 20$ and $b^{2} = 45$
$$LR = \frac{2a^{2}}{\text{ }b} = \frac{2(20)}{3\sqrt{5}} = \frac{8\sqrt{5}}{3} $$
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