The length of the latus rectum and directrice | vedclass

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(B) Given the length of latus rectum $\frac{2b^2}{a} = 9$ and directrices $x = \pm \frac{a}{e} = \pm \frac{4}{\sqrt{13}}$.From $\frac{a}{e} = \frac{4}

Vedclass · Accepted Answer

$61$

Vedclass · Answer

(B) Given the length of latus rectum $\frac{2b^2}{a} = 9$ and directrices $x = \pm \frac{a}{e} = \pm \frac{4}{\sqrt{13}}$.From $\frac{a}{e} = \frac{4}{\sqrt{13}}$,we have $a = \frac{4e}{\sqrt{13}}$.Also,$b^2 = a^2(e^2 - 1)$. Substituting $b^2 = \frac{9a}{2}$,we get $\frac{9a}{2} = a^2(e^2 - 1) \Rightarrow \frac{9}{2} = a(e^2 - 1) = \frac{4e}{\sqrt{13}}(e^2 - 1)$.Solving for $e$,we find $e = \frac{\sqrt{13}}{2}$ and $a = 2$. Then $b^2 = \frac{9(2)}{2} = 9$.The hyperbola is $\frac{x^2}{4} - \frac{y^2}{9} = 1$.The line $y = \sqrt{3}x - \sqrt{3}$ is a tangent. The condition of tangency $c^2 = a^2m^2 - b^2$ gives $(-\sqrt{3})^2 = 4(\sqrt{3})^2 - 9 = 12 - 9 = 3$,which holds.The point of contact $(x_0, y_0)$ is $(\frac{a^2m}{c}, \frac{b^2}{c}) = (\frac{4\sqrt{3}}{\sqrt{3}}, \frac{9}{\sqrt{3}}) = (4, 3\sqrt{3})$.The product of focal distances $m = e^2x_0^2 - a^2 = \frac{13}{4}(16) - 4 = 52 - 4 = 48$.Then $4e^2 + m = 4(\frac{13}{4}) + 48 = 13 + 48 = 61$.

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