Verified Solution • 2025–26 Session moderate

QJEEM21Ma4603: JEE Main 2021 Mathematics - Class 11 JEEM21Ma4603 Solution: A hyperbola passes through the foci of the ellipse and its…

A hyperbola passes through the foci of the ellipse x225+y216=1{{{x^2}} \over {25}} + {{{y^2}} \over {16}} = 1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is :

Multiple Choice Options

  • A x29−y24=1{{{x^2}} \over 9} - {{{y^2}} \over 4} = 1
  • B x29−y216=1{{{x^2}} \over 9} - {{{y^2}} \over 16} = 1
  • C x29−y225=1{{{x^2}} \over 9} - {{{y^2}} \over 25} = 1
  • D x2 −- y2 = 9

Step-by-Step Verified Solution

e1=1−1625=35{e_1} = \sqrt {1 - {{16} \over {25}}} = {3 \over 5} foci ( ±\pm ae, 0)

Foci = ( ±\pm 3, 0)

Let equation of hyperbola be x2A2−y2B2=1{{{x^2}} \over {{A^2}}} - {{{y^2}} \over {{B^2}}} = 1

Passes through ( ±\pm 3, 0)

A2 = 9, A = 3, e2=53{e_2} = {5 \over 3}

e22=1+B2A2{e_2}^2 = 1 + {{{B^2}} \over {{A^2}}}

259=1+B29⇒B2=16{{25} \over 9} = 1 + {{{B^2}} \over 9} \Rightarrow {B^2} = 16

Equation of the hyperbola

x29−y216=1{{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1

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