Question

Let any double ordinate $P N P'$ of the hyperbola $\frac{x^{2}}{25}-\frac{y^{2}}{16}=1$ be produced both sides to meet the asymptotes in $Q$ and $Q'$, then $P Q \cdot P' Q$ is equal to

Answer 1

$N P=\frac{4}{5}=\sqrt{x_{1}^{2}-25}$

$Q$ is on $y=\frac{4}{5} x$
$N Q=\frac{4}{5} x_{1}$
$P Q=N Q-N P=\frac{4}{5}\left(x_{1}+\sqrt{x_{1}^{2}-25}\right)$
$P' Q=\frac{4}{5}\left(x_{1}+\sqrt{x_{1}^{2}-25}\right)$
$P Q \cdot P' Q=16 .$