Verified Solution • 2025–26 Session moderate

QJEEM22Maf8f0: JEE Main 2022 Mathematics - Class 11 JEEM22Maf8f0 Solution: Let the eccentricity of an ellipse , , be . If this ellipse…

Let the eccentricity of an ellipse x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1 , a > b , be 14{1 \over 4} . If this ellipse passes through the point (−425,3)\left( { - 4\sqrt {{2 \over 5}} ,3} \right) , then a2+b2{a^2} + {b^2} is equal to :

Multiple Choice Options

  • A 29
  • B 31
  • C 32
  • D 34

Step-by-Step Verified Solution

x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1

⇒(−425)2a2+32b2=1\Rightarrow {{{{\left( { - 4\sqrt {{2 \over 5}} } \right)}^2}} \over {{a^2}}} + {{32} \over {{b^2}}} = 1

⇒325a2+9b2=1\Rightarrow {{32} \over {5{a^2}}} + {9 \over {{b^2}}} = 1 ..... (i)

a2(1−e2)=b2{a^2}(1 - {e^2}) = {b^2}

a2(1−116)=b2{a^2}\left( {1 - {1 \over {16}}} \right) = {b^2}

15a2=16b2⇒a2=16b21515{a^2} = 16{b^2} \Rightarrow {a^2} = {{16{b^2}} \over {15}}

From (i)

6b2+9b2=1⇒b2=15{6 \over {{b^2}}} + {9 \over {{b^2}}} = 1 \Rightarrow {b^2} = 15 & a2=16{a^2} = 16

a2+b2=15+16=31{a^2} + {b^2} = 15 + 16 = 31

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