Let LL- be the latus rectum through the focus S of a hyperbola and A - be the farther vertex of the conic-If - A - LL- is equilateral then its eccentricity e -A- -3-1-2B- -3C- -3-1D- -3-1-3

The correct option is D Given, A’LL’ is an equilatral triangle ∴∠ SA`L = 30^∘ tan 30^∘ = SL/A'S⇒1/√(3) = b^2 a/a(a+ae) ⇒1/√(3) = a^2(e^2 1)/(1+e)⇒ e = √(3) ...

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Standard XII

Physics

Relative Velocity

Let LL' be th...

Question

Let LL’ be the latus rectum through the focus S of a hyperbola and A’ be the farther vertex of the conic.If $Δ$ A’ LL’ is equilateral then its eccentricity e =

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Solution

The correct option is D
Given, A’LL’ is an equilatral triangle
$∴ ∠ S A ‘ L = 30^{\circ}$

$t a n 30^{\circ} = \frac{S L}{A^{'} S} \Rightarrow \frac{1}{\sqrt{3}} = \frac{b^{2} a}{a (a + a e)} \Rightarrow \frac{1}{\sqrt{3}} = \frac{a^{2} (e^{2} - 1)}{(1 + e)} \Rightarrow e = \frac{\sqrt{3} + 1}{\sqrt{3}}$

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Let LL’ be the latus rectum through the focus S of a hyperbola and A’ be the farther vertex of the conic.If Δ A’ LL’ is equilateral then its eccentricity e =

The correct option is D Given, A’LL’ is an equilatral triangle ∴∠SA‘L=30∘ tan 30∘=SLA′S⇒1√3=b2aa(a+ae)⇒1√3=a2(e2−1)(1+e)⇒e=√3+1√3

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