Let an ellipse x^2a^2+y^2b^2=1, a

Question

Let an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a<b$, pass through the point $(4,3)$ and have eccentricity $\dfrac{\sqrt5}{3}$. Then the length of its latus rectum is:

Accepted Answer

Since $a<b$, the major axis is along $y$, so $e^2=1-\dfrac{a^2}{b^2}=\dfrac59 \Rightarrow \dfrac{a^2}{b^2}=\dfrac49 \Rightarrow b^2=\dfrac94 a^2$.

Passing through $(4,3)$:

$$\frac{16}{a^2}+\frac{9}{b^2}=1 \;\Rightarrow\; \frac{16}{a^2}+\frac{9}{\tfrac94a^2}=1 \;\Rightarrow\; \frac{16}{a^2}+\frac{4}{a^2}=1 \;\Rightarrow\; a^2=20.$$

$$b^2=\frac94\times20=45.$$

Latus rectum (major axis along $y$) $=\dfrac{2a^2}{b}=\dfrac{40}{\sqrt{45}}=\dfrac{40}{3\sqrt5}=\dfrac{8\sqrt5}{3}$.

Solutions · 1