Question

How Wide Is an Ellipse at a Focus?

A latus rectum for an ellipse is a line segment perpendicular to the major axis at a focus, with endpoints on the ellipse, as shown in the figure. Show that the length of a latus rectum is $\frac{2b^2}{a}$ for the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$

Short answer

Hence proved that the latus rectum given as $\frac{2b^2}{a}$if for ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$

Step-by-step solution

Step 1. Given information

Equations of ellipse is given as-

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a>b$

Step 2. Concept used

A latus rectum for an ellipse is a line segment perpendicular to the major axis at a focus, with endpoints on the ellipse. Each fixed point is called a focus (plural: foci) of the ellipse.

In this problem, we have to prove that the latus rectum is $\frac{2b^2}{a}$given the equation of ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where$a>b$.

Step 3. Calculation

The endpoints of the latus rectum have the same $x$ coordinate as the foci.

Since the $x$ coordinate of the foci are

$\begin{array}{l}x=\pm c\\ x=\pm \sqrt{a^2-b^2}\end{array}$

Then,

$\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{{\left(\pm \sqrt{a^2-b^2}\right)}^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{a^2-b^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{a^2}{a^2}-\frac{b^2}{a^2}+\frac{y^2}{b^2}=1\\ 1-\frac{b^2}{a^2}+\frac{y^2}{b^2}=1\\ -\frac{b^2}{a^2}+\frac{y^2}{b^2}=0\\ \frac{y^2}{b^2}=\frac{b^2}{a^2}\\ y^2=\frac{b^4}{a^2}\\ y=\pm \sqrt{\frac{b^4}{a^2}}\\ y=\pm \frac{b^2}{a}\end{array}$

The length of the latus rectum is the difference of the y coordinates so it is

$\frac{b^2}{a}-\left(\frac{-b^2}{a}\right)=\frac{2b^2}{a}$

Hence proved