For some heta ≤ft(0, π/2), if the eccentricity of thehyperbola, x²-y² ² θ = 10 is {5} times theeccentricity of the ellipse, x² ² θ + y² = 5, thenthe length of the latus rectum of the ellipse, is: [2 Sep 2020 Shift 2]

Correct Answer is : 453{4 {5}}{3}345

Correct Answer is : 453{4 {5}}{3}345

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Hyperbola

Section: Mathematics

Question : 71 of 98

Marks: +1, -0

For some

heta \in \left(0, \frac{\pi}{2}\right),

x^{2}-y^{2} \sec^{2} \theta = 10

\sqrt{5}

x^{2} \sec^{2} \theta + y^{2} = 5,

Solution:

Given

\theta \in \left(0, \frac{\pi}{2}\right)

equation of hyperbola

\Rightarrow x^{2} - y^{2} \sec^{2} \theta = 10

\Rightarrow \frac{x^{2}}{10} - \frac{y^{2}}{10 \cos^{2} \theta} = 1

Hence eccentricity of hyperbola

(e_{H}) = \sqrt{1 + \frac{10 \cos^{2} \theta}{10}}

.....(1)

\{ e = \sqrt{1 + \frac{b^{2}}{a^{2}}} \}

Now equation of ellipse

\Rightarrow x^{2} \sec^{2} \theta + y^{2} = 5

\Rightarrow \frac{x^{2}}{5 \cos^{2} \theta} + \frac{y^{2}}{5} = 1

\{ e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \}

Hence eccenticity of ellipse

(e_{E}) = \sqrt{1 - \frac{5 \cos^{2} \theta}{5}}

(e_{E}) = \sqrt{1 - \cos^{2} \theta}

= |\sin \theta| = \sin \theta

…(2)

\{ \because \theta \in \left(0, \frac{\pi}{2}\right) \}

given

\Rightarrow e_{H} = \sqrt{5} e_{e}

Hence

1 + \cos^{2} \theta = 5 \sin^{2} \theta

1 + \cos^{2} \theta = 5 (1 - \cos^{2} \theta)

1 + \cos^{2} \theta = 5 - 5 \cos^{2} \theta

6 \cos^{2} \theta = 4

\cos^{2} \theta = \frac{2}{3}

…(3)

Now length of latus rectum of ellipse

= \frac{2 a^{2}}{b} = \frac{10 \cos^{2} \theta}{\sqrt{5}}

= \frac{20}{3 \sqrt{5}} = \frac{4 \sqrt{5}}{3}

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