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QJEEM24Ma84da: JEE Main 2024 Mathematics - Class 11 JEEM24Ma84da Solution: Let the line , intersect the -axis and -axis at the points and…

Let the line 2 x+3 y-\mathrm{k}=0, \mathrm{k}>0 , intersect the xx -axis and yy -axis at the points A\mathrm{A} and B\mathrm{B} , respectively. If the equation of the circle having the line segment ABA B as a diameter is x2+y2−3x−2y=0x^2+y^2-3 x-2 y=0 and the length of the latus rectum of the ellipse x2+9y2=k2x^2+9 y^2=k^2 is mn\frac{m}{n} , where mm and nn are coprime, then 2 m+n2 \mathrm{~m}+\mathrm{n} is equal to

Multiple Choice Options

  • A 12
  • B 13
  • C 11
  • D 10

Step-by-Step Verified Solution

JEE Main 2024 (Online) 5th April Morning Shift Mathematics - Ellipse Question 9 English Explanation

Equation of circle with ABA B as diameter

amp;(x−k2)x+y(y−k3)=0amp;⇒x2+y2−kx2−ky3=0\begin{aligned}& \left(x-\frac{k}{2}\right) x+y\left(y-\frac{k}{3}\right)=0 \\& \Rightarrow x^2+y^2-\frac{k x}{2}-\frac{k y}{3}=0\end{aligned}

Comparing, k=6k=6

Latus rectum of ellipse

amp;x2+9y2=k2=62amp;⇒x262+y222=1amp; L.R =2b2a=2×46=43amp;m=4amp;n=3amp;2m+n=8+3=11\begin{aligned}& x^2+9 y^2=k^2=6^2 \\& \Rightarrow \frac{x^2}{6^2}+\frac{y^2}{2^2}=1 \\& \text { L.R }=\frac{2 b^2}{a}=\frac{2 \times 4}{6}=\frac{4}{3} \\& m=4 \\& n=3 \\& 2 m+n=8+3=11\end{aligned}

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