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QJEEM19Ma0c50: JEE Main 2019 Mathematics - Class 11 JEEM19Ma0c50 Solution: If a directrix of a hyperbola centred at the origin and passing …

If a directrix of a hyperbola centred at the origin and passing through the point (4, –2 3\sqrt 3 ) is 5x = 4 5\sqrt 5 and its eccentricity is e, then :

Multiple Choice Options

  • A 4e4 – 24e2 + 27 = 0
  • B 4e4 – 24e2 + 35 = 0
  • C 4e4 – 12e2 - 27 = 0
  • D 4e4 + 8e2 - 35 = 0

Step-by-Step Verified Solution

5x = 4 5\sqrt 5

⇒\Rightarrow x=45x = {4 \over {\sqrt 5 }}

JEE Main 2019 (Online) 10th April Morning Slot Mathematics - Hyperbola Question 56 English Explanation
Equation of hyperbola

x2a2−y2b2=1{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1 it passes through (4,−23)\left( {4, - 2\sqrt 3 } \right)

∴e2=1+b2a2\therefore {e^2} = 1 + {{{b^2}} \over {{a^2}}}

⇒a2e2−a2=b2\Rightarrow {a^2}{e^2} - {a^2} = {b^2}

⇒16a2−12a2e2−a2=1\Rightarrow {{16} \over {{a^2}}} - {{12} \over {{a^2}{e^2} - {a^2}}} = 1

⇒4a2[41−3e2−1]=1\Rightarrow {4 \over {{a^2}}}\left[ {{4 \over 1} - {3 \over {{e^2} - 1}}} \right] = 1

⇒4e2−4−3=(e2−1)(a24)\Rightarrow 4{e^2} - 4 - 3 = ({e^2} - 1)\left( {{{{a^2}} \over 4}} \right)

⇒4(4e3−7)=(e2−1)(4e5)2\Rightarrow 4(4{e^3} - 7) = ({e^2} - 1){\left( {{{4e} \over {\sqrt 5 }}} \right)^2}

⇒4e4−24e2+35=0\Rightarrow 4{e^4} - 24{e^2} + 35 = 0

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