For some ( 0 , 2 ) , let the eccentricity and the length of the latus rectum of the hyperbola x 2 - y 2 sec 2 = 8 be e 1 a n d l 1 , respectively, and let the eccentricity and the length of the latus rectum of the ellipse

Question

For some $θ \in ((0, \frac{π}{2}))$ , let the eccentricity and the length of the latus rectum of the hyperbola $x^{2} - y^{2} \sec^{2} θ = 8$ be $e_{1} and l_{1},$ respectively, and let the eccentricity and the length of the latus rectum of the ellipse $x^{2} \sec^{2} θ + y^{2} = 6$ be $e_{2} and l_{2}$ , respectively. If $e_{1}^{2} = e_{2}^{2} ((s {ec}^{2} θ + 1)),$ then $((\frac{l_{1} l_{2}}{e_{1} e_{2}})) \tan^{2} θ$ is equal to______. [2026]

Smart Achievers · Accepted Answer

(8)

$\frac{x^{2}}{8} - \frac{y^{2}}{8 \cos^{2} θ} = 1, e_{1} = \sqrt{1 + \frac{8 \cos^{2} θ}{8}}$

$ℓ_{1} = \frac{2 b^{2}}{a} = \frac{2 (8 \cos^{2} θ)}{2 \sqrt{2}}$

$\frac{x^{2}}{6} + \frac{y^{2}}{6 \cos^{2} θ} = 1, e_{2} = \sqrt{1 - \frac{6 \cos^{2} θ}{6}} = \sin θ$

$ℓ_{2} = \frac{2 b^{2}}{a} = \frac{2 \cdot 6 \cos^{2} θ}{\sqrt{6}}$

$e_{1}^{2} = e_{2}^{2} (1 + \sec^{2} θ)$

$1 + \cos^{2} θ = \sin^{2} θ (1 + \frac{1}{\cos^{2} θ})$

$1 + \cos^{2} θ = \sin^{2} θ + \tan^{2} θ$

$Solving we get θ = \frac{π}{4}$

$ℓ_{1} = 2 \sqrt{2}$

$e_{1} = \sqrt{\frac{3}{2}}$

$ℓ_{2} = \sqrt{6}$

$e_{2} = \frac{1}{\sqrt{2}}$

$(\frac{ℓ_{1} ℓ_{2}}{e_{1} e_{2}}) \tan^{2} θ = 8 (By putting values)$

Solution

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Solution

Ans. (8) x28-y28cos2θ=1, e1=1+8cos2θ8 ℓ1=2b2a=2(8cos2θ)22 x26+y26cos2θ=1, e2=1-6cos2θ6=sinθ ℓ2=2b2a=2·6cos2θ6 e12=e22(1+sec2θ) 1+cos2θ=sin2θ(1+1cos2θ) 1+cos2θ=sin2θ+tan2θ Solving we get θ=π4 ℓ1=22 e1=32 ℓ2=6 e2=12 (ℓ1ℓ2e1e2)tan2θ=8 (By putting values)

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