Length of the Latus Rectum of an Ellipse
The length of the latus rectum of the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\,\,a > b$$ is $$\frac{{2{b^2}}}{a}$$.
The chord through the focus and perpendicular to the axis of the ellipse is called its latus rectum. Since the ellipse has two foci, it will have two latus recta.

Let $$A$$ and $$B$$ be the ends of the latus rectum as shown in the given diagram. Since the latus rectum passes through the focus, abscissa of $$A$$ and $$B$$ will be $$ae$$. Now putting $$x = ae$$ in the given equation of ellipse, we have
\[\begin{gathered} \frac{{{{\left( {ae} \right)}^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \\ \Rightarrow \frac{{{y^2}}}{{{b^2}}} = 1 – \frac{{{{\left( {ae} \right)}^2}}}{{{a^2}}} \\ \Rightarrow \frac{{{y^2}}}{{{b^2}}} = \frac{{{a^2} – {{\left( {ae} \right)}^2}}}{{{a^2}}}\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered} \]
Since $${a^2} – {\left( {ae} \right)^2} = {b^2}$$, putting this value in equation (i), we have
\[\frac{{{y^2}}}{{{b^2}}} = \frac{{{b^2}}}{{{a^2}}} \Rightarrow {y^2} = \frac{{{b^4}}}{{{a^2}}} \Rightarrow y = \pm \frac{{{b^2}}}{a}\]
Thus, $$A\left( {ae,\frac{{{b^2}}}{a}} \right)$$ and $$B\left( {ae, – \frac{{{b^2}}}{a}} \right)$$. The length of the latus rectum is
\[\begin{gathered} l = \left| {AB} \right| = \sqrt {{{\left( {ae – ae} \right)}^2} + {{\left( {\frac{{{b^2}}}{a} – \left( { – \frac{{{b^2}}}{a}} \right)} \right)}^2}} \\ \Rightarrow l = \sqrt {{{\left( {\frac{{2{b^2}}}{a}} \right)}^2}} = \frac{{2{b^2}}}{a} \\ \end{gathered} \]
Anuj Jalwal
May 24 @ 9:55 pm
How did you know about
a^2-(ae)^2 = b^2
Phil
April 1 @ 1:26 pm
ae (sometimes termed as “c”) is the location of the foci from the center
the formula is
c^2=a^2 – b^2
thats where we can also get a^2 – c^2 = b^2