The length of the latus rectum and directrices of hyperbola with eccentricity e are 9 and …

Question

The length of the latus rectum and directrices of hyperbola with eccentricity e are 9 and $$x= \pm \frac{4}{\sqrt{3}}$$, respectively. Let the line $$y-\sqrt{3} x+\sqrt{3}=0$$ touch this hyperbola at $$\left(x_0, y_0\right)$$

Accepted Answer

$y=mx\pm \sqrt{{a}^{2}{m}^{2}-{b}^{2}}$ is the tangent

$\begin{gathered} \Rightarrow m=\sqrt{3}\Rightarrow {a}^{2}{m}^{2}-{b}^{2}=3 \ \Rightarrow 3{a}^{2}-{b}^{2}=3 \end{gathered}$

$\begin{gathered} \frac{2{b}^{2}}{a}=9\Rightarrow {b}^{2}=\frac{9a}{2}\Rightarrow 3{a}^{2}-\frac{9a}{2}=3 \ \Rightarrow {a}^{2}-\frac{3}{2}a-1=0 \ \Rightarrow a=2-0.5 \ \Rightarrow b=3 \ \Rightarrow \frac{{x}^{2}}{4}-\frac{{y}^{2}}{9}=1 \ \Rightarrow y=\sqrt{3x}-\sqrt{3} \ {x}_{0}=4,{y}_{0}=3\sqrt{3},e=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2} \ P{F}_{1}\cdot P{F}_{2}= \ \sqrt{(4-\sqrt{13}{)}^{2}+(3\sqrt{3}{)}^{2}}\sqrt{(4+\sqrt{13}{)}^{2}+(3\sqrt{3}{)}^{2}} \ =\sqrt{(56-8\sqrt{13})(56+8\sqrt{13})} \ =\sqrt{2304}=48=m \ \Rightarrow 4{e}^{2}+m=13+48=61 \end{gathered}$

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