Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b,…
Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be $(-5,0)$ and $5 x+9=0$, respectively. If the product of the focal distances of a point $(\alpha, 2 \sqrt{5})$ on the hyperbola is $p$, then $4 p$ is equal to
Solution
Equation of hyperbola is \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Directrix: \(\mathrm{x}=\frac{-9}{5}\) and corresponding foci \((-5,0)\)
\(\Rightarrow-\frac{\mathrm{a}}{\mathrm{e}}=-\frac{9}{5}\) and \(-\mathrm{ae}=-5\)
\(\Rightarrow \frac{9 e^2}{5}=5 \Rightarrow e=\sqrt{\frac{25}{9}}=\frac{5}{3} \Rightarrow a=3\)
\(\therefore \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)=9\left(\frac{25}{9}-1\right)=16\)
Hyperbola \(\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1\)
\((\alpha, 2 \sqrt{5})\) lie on it
\(\Rightarrow \frac{\alpha^2}{9}-\frac{20}{16}=1 \Rightarrow \alpha^2=\frac{36}{16} \times 9=\frac{81}{4}\)
Product for distance of \(\left(\mathrm{x}_1 \mathrm{y}_1\right)\) from the two foci
\(\begin{aligned}
& =\left(e x_1+a\right)\left|e x_1-a\right| \\
& =e^2 x_1^2-a^2
\end{aligned}\)
For \((\alpha, 2 \sqrt{5}) \Rightarrow \mathrm{P}=\frac{25}{9} \cdot \frac{81}{4}-9=\frac{189}{4}\)
\(4 \mathrm{P}=189\)