Verified Solution • 2025–26 Session moderate

QJEEM24Ma09eb: JEE Main 2024 Mathematics - Class 11 JEEM24Ma09eb Solution: Let and . If and denote the eccentricity and the length of the…

Let f(x)=x2+9,g(x)=xx−9f(x)=x^2+9, g(x)=\frac{x}{x-9} and a=f∘g(10),b=g∘f(3)\mathrm{a}=f \circ g(10), \mathrm{b}=g \circ f(3) . If e\mathrm{e} and ll denote the eccentricity and the length of the latus rectum of the ellipse x2a+y2 b=1\frac{x^2}{\mathrm{a}}+\frac{y^2}{\mathrm{~b}}=1 , then 8e2+l28 \mathrm{e}^2+l^2 is equal to.

Multiple Choice Options

  • A 6
  • B 12
  • C 8
  • D 16

Step-by-Step Verified Solution

amp;g(10)=10amp;a=f(g(10))=f(10)=109amp;f(3)=18amp;b=g(f(3))=g(18)=2amp;x2109+y22=1amp;e=1−2109=107109amp;I=2b2a=2×2109\begin{aligned}& g(10)=10 \\& a=f(g(10))=f(10)=109 \\& f(3)=18 \\& b=g(f(3))=g(18)=2 \\& \frac{x^2}{109}+\frac{y^2}{2}=1 \\& e=\sqrt{1-\frac{2}{109}}=\sqrt{\frac{107}{109}} \\& I=\frac{2 b^2}{a}=\frac{2 \times 2}{\sqrt{109}}\end{aligned}

8e2+l2amp;=8×107109+16109amp;=8\begin{aligned}8 e^2+l^2 & =\frac{8 \times 107}{109}+\frac{16}{109} \\& =8\end{aligned}

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