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JEE Main 2019 · 10 April · Shift I

JEE Main 2019 Mathematics question on Conic Sections

If a directrix of a hyperbola centred at the origin and passing through the point (4,−23)(4, -2\sqrt{3}) is 5x=455x = 4\sqrt{5} and its eccentricity is ee, then
  1. A.
    4e4−12e2−27=04e^4 - 12e^2 - 27 = 0
  2. B.
    4e4−24e2+27=04e^4 - 24e^2 + 27 = 0
  3. C.
    4e4+8e2−35=04e^4 + 8e^2 - 35 = 0
  4. D.
    4e4−24e2+35=04e^4 - 24e^2 + 35 = 0
Show answer and solution

Correct option: D

Solution

The directrix is x = 4√5/5, so a/e = 4√5/5, giving a = 4e√5/5.

Since the hyperbola passes through (4, -2√3), using x²/a² - y²/b² = 1 with b² = a²(e²-1):

16/a² - 12/[a²(e²-1)] = 1

Substituting a² = 80e²/25 = 16e²/5:

16/(16e²/5) - 12/[(16e²/5)(e²-1)] = 1

5/e² - 15/[4e²(e²-1)] = 1

Multiply through by 4e²(e²-1):

20(e²-1) - 15 = 4e²(e²-1)

20e² - 20 - 15 = 4e⁴ - 4e²

20e² - 35 = 4e⁴ - 4e²

4e⁴ - 24e² + 35 = 0

This matches option 4.

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