Question

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

• A. $25$
• B. $16$
• C. $41$
• D. None of these

Answer 1

The correct option is B $16$

Let $P(x_1,y_1)$ and $P^{\prime }(x_1,-y_1)$ be the double ordinate of the hyperbola.
$\Rightarrow \frac{{x_1}^2}{25}-\frac{y_1^2}{16}=1$
$\Rightarrow y_1=\frac{4}{5}\sqrt{x_1^2-25}$
$NP=\frac{4}{5}\sqrt{x_1^2-25}$
Since $Q$ lies on the asymptote of the hyperbola
$Q$ is on $y=\frac{4}{5}x$
$\Rightarrow NQ=\frac{4}{5}{x}_1$
$PQ=NQ-NP=\frac{4}{5}(x_1-\sqrt{x_1^2-25})$
$P^{\prime }Q=NQ+N{P}^{\prime }=\frac{4}{5}(x_1+\sqrt{x_1^2-25})$
$PQ\cdot P^{\prime }Q=16$