Let the sum of the focal distances of the point P(4,3) on the hyperbola H:x^2a^2-y^2b^2=1 …

Question

Let the sum of the focal distances of the point $P(4,3)$ on the hyperbola $H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ 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 $9{l}^{2}+6m$ is equal to :

Accepted Answer

$\begin{gathered} \sqrt{(c+4{)}^{2}+9}+\sqrt{(c-4{)}^{2}+9}=8\sqrt{\frac{5}{3}} \ c=\frac{5}{\sqrt{6}}=al\Rightarrow {a}^{2}(1+\frac{{b}^{2}}{{a}^{2}})=\frac{25}{6}\Rightarrow {a}^{2}+{b}^{2}=\frac{25}{6} \ \frac{16}{{a}^{2}}-\frac{9}{{b}^{2}}=1 \ 16{b}^{2}-9{d}^{2}={a}^{2}{b}^{2}\Rightarrow 16(\frac{25}{6}-{a}^{2})-9{a}^{2}=9{a}^{2}{b}^{2} \ P{F}_{1}+P{F}_{2}=8\sqrt{\frac{5}{3}}\Rightarrow {a}^{2}=\frac{5}{2},{b}^{2}=\frac{5}{3} \ |P{F}_{1}-P{F}_{2}|=2a \ \frac{64.5}{3}=4{a}^{2}+4m\Rightarrow m=\frac{80}{3}-{a}^{2} \ 6m=160-6{a}^{2} \end{gathered}$

$\begin{gathered} 9{l}^{2}=9{(\frac{2{b}^{2}}{a})}^{2}=\frac{36{b}^{4}}{{a}^{2}} \ 9{l}^{2}+6m=\frac{36(\frac{25}{9})}{\frac{5}{2}}+160-6(\frac{5}{2}) \ =\frac{72	imes 5}{9}+160-15 \ =160+40-15=185 \end{gathered}$

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