Verified Solution • 2025–26 Session moderate

QJEEM22Ma281c: JEE Main 2022 Mathematics - Class 11 JEEM22Ma281c Solution: Let the hyperbola pass through the point . A parabola is drawn…

Let the hyperbola H:x2a2−y2b2=1H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 pass through the point (22,−22)(2 \sqrt{2},-2 \sqrt{2}) . A parabola is drawn whose focus is same as the focus of H\mathrm{H} with positive abscissa and the directrix of the parabola passes through the other focus of H\mathrm{H} . If the length of the latus rectum of the parabola is e times the length of the latus rectum of H\mathrm{H} , where e is the eccentricity of H, then which of the following points lies on the parabola?

Multiple Choice Options

  • A (23,32)(2 \sqrt{3}, 3 \sqrt{2})
  • B (33,−62)\mathbf(3 \sqrt{3},-6 \sqrt{2})
  • C (3,−6)(\sqrt{3},-\sqrt{6})
  • D (36,62)(3 \sqrt{6}, 6 \sqrt{2})

Step-by-Step Verified Solution

H:x2a2−y2b2=1H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1

Focus of parabola : (ae, 0)(ae,\,0)

Directrix : x=−aex = - ae .

Equation of parabola ≡y2=4aex\equiv {y^2} = 4aex

Length of latus rectum of parabola =4ae= 4ae

Length of latus rectum of hyperbola =2.b2a= {{2.{b^2}} \over a}

as given, 4ae=2b2a . e4ae = {{2{b^2}} \over a}\,.\,e

2=b2a22 = {{{b^2}} \over {{a^2}}} ...... (i)

∵\because H passes through (22,−22)⇒8a2−8b2=1\left( {2\sqrt 2 , - 2\sqrt 2 } \right) \Rightarrow {8 \over {{a^2}}} - {8 \over {{b^2}}} = 1 ........ (ii)

From (i) and (ii) a2=4{a^2} = 4 and b2=8⇒e=3{b^2} = 8 \Rightarrow e = \sqrt 3

⇒\Rightarrow Equation of parabola is y2=83x{y^2} = 8\sqrt 3 x .

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