Let the product of the focal distances of the point P(4, 2√3) on the hyperbola H: x^2/a^2 - y^2/b^2 = 1 be 32.

Question

Let the product of the focal distances of the point \(\mathrm{P}(4,2 \sqrt{3})\) on the hyperbola \(\mathrm{H}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\) be 32 . Let the length of the conjugate axis of H be p and the length of its latus rectum be q . Then \(\mathrm{p}^{2}+\mathrm{q}^{2}\) is equal to ......

Answer 1

Answer is: 120 

\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \quad \cdots(1)\)

\(\mathrm{P}(4,2 \sqrt{3})\)

\(\mathrm{PS}_{1} \cdot \mathrm{PS}_{2}=32\)

\(\left|\mathrm{PS}_{1}-\mathrm{PS}_{2}\right|=2 \mathrm{a}\)

\(\mathrm{P}(4,2 \sqrt{3})\) lies on H

\(\therefore \frac{16}{\mathrm{a}^{2}}-\frac{12}{\mathrm{~b}^{2}}=1\)

\(16 b^{2}-12 a^{2}=a^{2} b^{2}\quad \cdots(2)\)

\(\left|\mathrm{PS}_{1}-\mathrm{PS}_{2}\right|^{2}=4 \mathrm{a}^{2}\)

\(\mathrm{PS}_{1}{ }^{2}+\mathrm{PS}_{2}^{2}-2 \mathrm{PS}_{1} \cdot \mathrm{PS}_{2}=4 \mathrm{a}^{2}\)

\((a e-4)^{2}+12+(a e+4)^{2}+12-64=4 a^{2}\)

\(2 \mathrm{a}^{2} \mathrm{e}^{2}-8=4 \mathrm{a}^{2}\)

\(a^{2}+b^{2}-4=2 a^{2}\)

\(b^{2}-a^{2}=4\)

\((2)\ \& \ (3) \Rightarrow 16\left(a^{2}+4\right)-12 a^{2}=a^{2}\left(a^{2}+4\right)\)

\(\Rightarrow 16 a^{2}+64-12 a^{2}=a^{4}+4 a^{2}\)

\(\Rightarrow a^{4}=64\)

\(\Rightarrow \mathrm{a}^{2}=8\)

\(\therefore \mathrm{b}^{2}=12\)

\(p^{2}+q^{2}=4 b^{2}+\frac{4 b^{4}}{a^{2}}\)

= 120