Let the sum of the focal distances of the point P(4,3) on the hyperbola H: x^2a^2-y^2~b^2=…

Question

Let the sum of the focal distances of the point $\mathrm{P}(4,3)$ on the hyperbola $\mathrm{H}: \frac{x^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}=1$ be $8 \sqrt{\frac{5}{3}}$. If for H , the length of the latus rectum is $l$ and the product of the focal distances of the point P is m , then $9 l^2+6 \mathrm{~m}$ is equal to :

Accepted Answer

$$\begin{aligned}
&\sqrt{(c+4)^2+9}+\sqrt{(c-4)^2+9}=8 \sqrt{\frac{5}{3}}\
&\text { Solving, } c=\frac{5}{\sqrt{6}}=\mathrm{al} \Rightarrow a^2\left(1+\frac{b^2}{a^2}\right)=\frac{25}{6} \Rightarrow a^2+b^2=\frac{25}{6}\
&\begin{aligned}
& \frac{16}{a^2}-\frac{9}{b^2}=1 \
& 16 b^2-9 d^2=a^2 b^2 \Rightarrow 16\left(\frac{25}{6}-a^2\right)-9 a^2=9 a^2 b^2 \
& P F_1+P F_2=8 \sqrt{\frac{5}{3}} \Rightarrow a^2=\frac{5}{2}, b^2=\frac{5}{3} \
& \left|P F_1-P F_2\right|=2 a \
& \frac{64.5}{3}=4 a^2+4 m \Rightarrow m=\frac{80}{3}-a^2 \
& 6 m=160-6 a^2
\end{aligned}
\end{aligned}$$

$$\begin{aligned}
& 9 \ell^2=9\left(\frac{2 b^2}{a}\right)^2=\frac{36 b^4}{a^2} \
& 9 \ell^2+6 m=\frac{36\left(\frac{25}{9}\right)}{\frac{5}{2}}+160-6\left(\frac{5}{2}\right) \
& =\frac{72 \times 5}{9}+160-15 \
& =160+40-15=185
\end{aligned}$$

Solutions · 1