Question

Sol. [Key Concept: Solve $6e^{2} - 11e + 3 = 0$ (reject $e < 1$); $2ae = 9$ gives $a$, and the latus rectum $= \frac{2b^{2}}{a} = 2a(e^{2} - 1)$.]

$$S_{1}(3,5)\&{\text{ }S}_{2}(3, - 4) $$$\because S_{1}{\text{ }S}_{2} = 2ae$ $\begin{array}{r} \Rightarrow 9 = 2ae\#(1) \end{array}$

and $\because 6e^{2} - 11e + 3 = 0$

$\Rightarrow e = \frac{3}{2},e = \frac{1}{3} < 1$ (rejected)

From (1); $a = 3$

$$LR = 2 \cdot \frac{{\text{ }b}^{2}}{a} = 2a\left( e^{2} - 1 \right) = 2.3\left( \frac{9}{4} - 1 \right) = \frac{15}{2} $$