Verified Solution • 2025–26 Session moderate

QJEEM24Mab488: JEE Main 2024 Mathematics - Class 11 JEEM24Mab488 Solution: Let be a parabola with vertex and directrix . Let an ellipse ,…

Let PP be a parabola with vertex (2,3)(2,3) and directrix 2x+y=62 x+y=6 . Let an ellipse E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b , of eccentricity 12\frac{1}{\sqrt{2}} pass through the focus of the parabola PP . Then, the square of the length of the latus rectum of EE , is

Multiple Choice Options

  • A 51225\frac{512}{25}
  • B 65625\frac{656}{25}
  • C 3858\frac{385}{8}
  • D 3478\frac{347}{8}

Step-by-Step Verified Solution

JEE Main 2024 (Online) 31st January Evening Shift Mathematics - Ellipse Question 13 English Explanation

amp; Slope of axis =12amp;y−3=12(x−2)amp;⇒2y−6=x−2amp;⇒2y−x−4=0amp;2x+y−6=0amp;4x+2y−12=0amp;α+1.6=4⇒α=2.4amp;β+2.8=6⇒β=3.2\begin{aligned}& \text { Slope of axis }=\frac{1}{2} \\& y-3=\frac{1}{2}(x-2) \\& \Rightarrow 2 y-6=x-2 \\& \Rightarrow 2 y-x-4=0 \\& 2 x+y-6=0 \\& 4 x+2 y-12=0 \\& \alpha+1.6=4 \Rightarrow \alpha=2.4 \\& \beta+2.8=6 \Rightarrow \beta=3.2\end{aligned}

Ellipse passes through (2.4,3.2)(2.4,3.2)

⇒(2410)2a2+(3210)2 b2=1\Rightarrow \frac{\left(\frac{24}{10}\right)^2}{\mathrm{a}^2}+\frac{\left(\frac{32}{10}\right)^2}{\mathrm{~b}^2}=1 ..... (1)

Also 1−b2a2=12=b2a2=121-\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{2}=\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1}{2}

⇒a2=2b2\Rightarrow a^2=2 b^2

Put in (1) ⇒b2=32825\Rightarrow b^2=\frac{328}{25}

⇒(2 b2a)2=4 b2a2×b2=4×12×32825=65625\Rightarrow\left(\frac{2 \mathrm{~b}^2}{\mathrm{a}}\right)^2=\frac{4 \mathrm{~b}^2}{\mathrm{a}^2} \times \mathrm{b}^2=4 \times \frac{1}{2} \times \frac{328}{25}=\frac{656}{25}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring