Let the latus rectum of the hyperbola frac{x^ | vedclass

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(C) The latus rectum of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passes through the focus $(ae, 0)$ and is perpendicular to the transverse ax

Vedclass · Accepted Answer

$182$

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(C) The latus rectum of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passes through the focus $(ae, 0)$ and is perpendicular to the transverse axis. The endpoints of the latus rectum are $(ae, \frac{b^2}{a})$ and $(ae, -\frac{b^2}{a})$.The angle subtended by the latus rectum at the centre $(0, 0)$ is $\frac{\pi}{3} = 60^{\circ}$.Considering the right-angled triangle formed by the centre,the focus,and one endpoint of the latus rectum,the angle at the centre is $\frac{60^{\circ}}{2} = 30^{\circ}$.Thus,$	an 30^{\circ} = \frac{b^2/a}{ae} = \frac{b^2}{a^2e} = \frac{1}{\sqrt{3}}$.Given $a^2 = 9$,we have $a = 3$. So,$\frac{b^2}{9e} = \frac{1}{\sqrt{3}} \Rightarrow b^2 = 3\sqrt{3}e$.We know $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{b^2}{9} \Rightarrow b^2 = 9(e^2 - 1)$.Equating the two expressions for $b^2$: $9(e^2 - 1) = 3\sqrt{3}e$ $\Rightarrow 3(e^2 - 1) = \sqrt{3}e$ $\Rightarrow 3e^2 - \sqrt{3}e - 3 = 0$.Solving for $e$: $e = \frac{\sqrt{3} \pm \sqrt{3 - 4(3)(-3)}}{2(3)} = \frac{\sqrt{3} + \sqrt{39}}{6} = \frac{\sqrt{3}(1 + \sqrt{13})}{6} = \frac{1 + \sqrt{13}}{2\sqrt{3}}$.Then $b^2 = 3\sqrt{3} \left( \frac{1 + \sqrt{13}}{2\sqrt{3}} ight) = \frac{3}{2}(1 + \sqrt{13})$.Comparing with $\frac{l}{m}(1+\sqrt{n})$,we get $l=3, m=2, n=13$.Therefore,$l^2+m^2+n^2 = 3^2 + 2^2 + 13^2 = 9 + 4 + 169 = 182$.

Let the latus rectum of the hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$ subtend an angle of $\frac{\pi}{3}$ at the centre of the hyperbola. If $b^2$ is equal to $\frac{l}{m}(1+\sqrt{n})$,where $l$ and $m$ are co-prime numbers,then $l^2+m^2+n^2$ is equal to . . . . . . .

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