One of the latus recta of the hyperbola frac{ | vedclass

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(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The coordinates of the endpoints of the latus rectum are $(ae, b^2/a)$ a

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$4$

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(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The coordinates of the endpoints of the latus rectum are $(ae, b^2/a)$ and $(ae, -b^2/a)$.Let the angle subtended by the latus rectum at the centre $(0,0)$ be $2	heta$. Then $	an 	heta = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$.Given $2	heta = 2 \operatorname{Tan}^{-1}\left(\frac{3}{2}ight)$,so $	an 	heta = \frac{3}{2}$.Thus,$\frac{b^2}{a^2e} = \frac{3}{2}$.Given $b^2 = 36$,we have $\frac{36}{a^2e} = \frac{3}{2}$,which implies $a^2e = 24$.We know $b^2 = a^2(e^2 - 1)$,so $36 = a^2e^2 - a^2$. Since $a^2e = 24$,$a^2 = \frac{24}{e}$.Substituting $a^2$,we get $36 = (\frac{24}{e})e^2 - \frac{24}{e} \implies 36 = 24e - \frac{24}{e}$.Dividing by $12$,$3 = 2e - \frac{2}{e} \implies 2e^2 - 3e - 2 = 0$.Solving the quadratic equation,$(2e+1)(e-2) = 0$. Since $e > 1$,$e = 2$.Then $a^2 = \frac{24}{2} = 12$.Finally,$\sqrt{a^2 + e^2} = \sqrt{12 + 2^2} = \sqrt{12 + 4} = \sqrt{16} = 4$.

One of the latus recta of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends an angle $2 \operatorname{Tan}^{-1}\left(\frac{3}{2}\right)$ at the centre of the hyperbola. If $b^2=36$ and $e$ is the eccentricity of the given hyperbola,then $\sqrt{a^2+e^2}=$

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