Answer
Length of the latera recta: $\frac{2b^2}{a}$
Work Step by Step
Use:
$a^2=b^2+c^2$
$a^2-c^2=b^2$
Standard form when major axis is horizontal:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Let's find the values of $y$ when $x=±c$:
$\frac{(±c)^2}{a^2}+\frac{y^2}{b^2}=1~~$
$\frac{y^2}{b^2}=1-\frac{c^2}{a^2}$
$\frac{y^2}{b^2}=\frac{a^2-c^2}{a^2}$
$\frac{y^2}{b^2}=\frac{b^2}{a^2}$
$y^2=\frac{b^4}{a^2}$
$y=±\frac{b^2}{a}$
One of the latera rectas, the one on the right side, goes from $(c,-\frac{b^2}{a})$ to $(c,\frac{b^2}{a})$. So, the length is: $\frac{b^2}{a}-(-\frac{b^2}{a})=\frac{2b^2}{a}$
