Latus Rectum - Practice Questions & MCQ

Length of Latus rectum

Let Latus rectum $\mathrm{LL}^{\prime}=2 \alpha$
$\mathrm{S}(\mathrm{ae}, 0)$ is focus, then $\mathrm{LS}=\mathrm{SL}^{\prime}=\alpha$
Coordinates of L and $\mathrm{L}^{\prime}$ become ( $\mathrm{ae}, \alpha$ ) and (ae, $-\alpha$ ) respectively
Equation of ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Put $x=a e, y=\alpha$ we get, $\frac{(\mathrm{ae})^2}{\mathrm{a}^2}+\frac{\alpha^2}{\mathrm{~b}^2}=1 \Rightarrow \alpha^2=\mathrm{b}^2\left(1-\mathrm{e}^2\right)$

$
\begin{aligned}
& \alpha^2=\mathrm{b}^2\left(\frac{\mathrm{~b}^2}{\mathrm{a}^2}\right) \quad\left[\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)\right] \\
& \alpha=\frac{\mathrm{b}^2}{\mathrm{a}} \\
& \Rightarrow 2 \alpha=\mathrm{LL}^{\prime}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}
\end{aligned}
$

End Points of Latus rectum

$\mathrm{L}=\left(\mathrm{ae}, \frac{\mathrm{v}}{\mathrm{a}}\right)$ and $\mathrm{L}^{\prime}=\left(\mathrm{ae},-\frac{\mathrm{v}}{\mathrm{a}}\right)$

Focal Distance of a Point:

The sum of the focal distance of any point on the ellipse is equal to the major axis.

Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be any point on the ellipse.

Here, $\quad \begin{aligned} & S P=e P M=e\left(\frac{a}{e}-x\right)=a-e x \\ & \\ & S^{\prime} P=e P M^{\prime}=e\left(\frac{a}{e}+x\right)=a+e x\end{aligned}$

Now, $\mathrm{SP}+\mathrm{S}^{\prime} \mathrm{P}=\mathrm{a}-\mathrm{ex}+\mathrm{a}+\mathrm{ex}=2 \mathrm{a}=\mathrm{AA}^{\prime}=$ constant.
Thus the sum of the focal distances of a point on the ellipse is constant.