Question

The vertices of the triangle $ABC$ are $A\left(0,0\right),B\left(3,0\right)$ and $C\left(3,4\right)$ , where $A$ and $C$ are foci of an ellipse and $B$ lies on the ellipse. If the length of the latus rectum of the ellipse is $\frac{12}{p}$ units, then the value of $p$ is

Answer 1

Answer: 3.5

$AC=5=2ae$
$\Rightarrow 2ae=5\dots \left(i\right)$
Also $AB+BC=2a$
$\Rightarrow 2a=7$ $\Rightarrow a=\frac{7}{2}$
$\therefore e=\frac{5}{7}$
$\Rightarrow 1-\frac{b^2}{a^2}=\frac{25}{49}\Rightarrow b=\sqrt{6}$
Length of the latus rectum, $\frac{2b^2}{a}=\frac{2\times 6}{\frac{7}{2}}$ units
$\Rightarrow p=\frac{7}{2}=3.5$