Question ID: #936
For some $\theta\in(0,\frac{\pi}{2}),$ let the eccentricity and the length of the latus rectum of the hyperbola $x^{2}-y^{2}\sec^{2}\theta=8$ be $e_{1}$ and $l_{1}$, respectively, and let the eccentricity and the length of the latus rectum of the ellipse $x^{2}\sec^{2}\theta+y^{2}=6$ be $e_{2}$ and $l_{2},$ respectively. If $e_{1}^{2}=e_{2}^{2}(\sec^{2}\theta+1)$, then $\left(\frac{l_{1}l_{2}}{e_{1}e_{2}}\right)\tan^{2}\theta$ is equal to
Solution:
**Hyperbola:** $x^2 – y^2\sec^2\theta = 8 \Rightarrow \frac{x^2}{8} – \frac{y^2}{8\cos^2\theta} = 1$.
Here $a^2 = 8, b^2 = 8\cos^2\theta$.
$$e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \cos^2\theta}$$
$$l_1 = \frac{2b^2}{a} = \frac{2(8\cos^2\theta)}{\sqrt{8}} = \frac{16\cos^2\theta}{2\sqrt{2}} = 4\sqrt{2}\cos^2\theta$$
**Ellipse:** $x^2\sec^2\theta + y^2 = 6 \Rightarrow \frac{x^2}{6\cos^2\theta} + \frac{y^2}{6} = 1$.
Since $\theta \in (0, \pi/2)$, $\cos^2\theta < 1$, so $6\cos^2\theta < 6$. Thus, major axis is along y-axis with $B^2 = 6, A^2 = 6\cos^2\theta$. $$e_2 = \sqrt{1 - \frac{A^2}{B^2}} = \sqrt{1 - \cos^2\theta} = \sin\theta$$ $$l_2 = \frac{2A^2}{B} = \frac{2(6\cos^2\theta)}{\sqrt{6}} = 2\sqrt{6}\cos^2\theta$$
**Condition:** $e_1^2 = e_2^2(\sec^2\theta + 1)$
$$1 + \cos^2\theta = \sin^2\theta \left( \frac{1}{\cos^2\theta} + 1 \right)$$
$$1 + \cos^2\theta = \sin^2\theta \left( \frac{1+\cos^2\theta}{\cos^2\theta} \right)$$
$$1 = \frac{\sin^2\theta}{\cos^2\theta} \Rightarrow \tan^2\theta = 1 \Rightarrow \theta = \frac{\pi}{4}$$
**Calculate Values at $\theta = \pi/4$:**
$$e_1 = \sqrt{1 + 1/2} = \sqrt{\frac{3}{2}}, \quad e_2 = \frac{1}{\sqrt{2}}$$
$$l_1 = 4\sqrt{2}\left(\frac{1}{2}\right) = 2\sqrt{2}, \quad l_2 = 2\sqrt{6}\left(\frac{1}{2}\right) = \sqrt{6}$$
**Required Expression:**
$$\left(\frac{l_1 l_2}{e_1 e_2}\right)\tan^2\theta = \frac{(2\sqrt{2})(\sqrt{6})}{(\sqrt{3/2})(1/\sqrt{2})} \cdot (1)$$
$$= \frac{2\sqrt{12}}{\sqrt{3}/2} = \frac{4\sqrt{3}}{\sqrt{3}/2} = 8$$
Ans. 8
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