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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Hyperbola

NTA Abhyas 2020 · Mathematics · Hyperbola

A hyperbola has foci (4,2), (2, 2) and it passes throughP(2, 4) . &...

A hyperbola has foci (4,2),(2,2)\left(4,2\right), \left(2, 2\right) and it passes throughP\left(2, 4\right)..& The eccentricity of the hyperbola is

Options

  1. Atan3π10tan \dfrac{3\pi}{10}
  2. Btan5π12tan \dfrac{5\pi}{12}
  3. Ctanπ3tan \dfrac{\pi}{3}
  4. Dtan3π8tan \dfrac{3\pi}{8}

Correct answer

D. tan3π8tan \dfrac{3\pi}{8}

Step-by-step solution

Let (4,2)=S1\left(4, 2\right) = {S}_{1} and \left(2, 2\right) = {S}_{2}and</p><p>theeccentricityofthehyperbolaisand</p><p>the eccentricity of the hyperbola ise,then</p><p>S1S2=2aeand∣PS1−PS2∣=2a, then</p><p>{S}_{1}{S}_{2} = 2ae and \left|P{S}_{1} - P{S}_{2}\right| = 2a (where,~2a2a is the length of the transverse axis)
⇒e=S1S2∣PS1−PS2∣=222−2⇒ e = \dfrac{{S}_{1}{S}_{2}}{\left|P{S}_{1} - P{S}_{2}\right|} = \dfrac{2}{2\sqrt{2} - 2}
⇒e=12−1=2+1=tan3π8⇒ e = \dfrac{1}{\sqrt{2} - 1} = \sqrt{2} + 1 = tan \dfrac{3\pi}{8}

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