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JEE Main 20245 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Ellipse

Let the line 2 x+3 y-k=0, k>0 , intersect the x -axis and y -axis...

Let the line 2x+3y−k=0,k>02 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 AB\mathrm{AB} 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

Options

  1. A11
  2. B10
  3. C12
  4. D13

Correct answer

A. 11

Step-by-step solution

Centre of the circle =\left(\frac{3}{2}, 1\right) Equation of diameter =2=2\mathrm{x}+3 \mathrm{y}-\mathrm{k}=0 2(32)+3(1)−k=0⇒k=6\begin{aligned} & 2\left(\frac{3}{2}\right)+3(1)-\mathrm{k}=0 \\ & \Rightarrow \mathrm{k}=6 \end{aligned}Now,EquationofellipsebecomesNow, Equation of ellipse becomesx2+9y2=36x262+y222=1\begin{aligned} & x^2+9 y^2=36 \\ & \frac{x^2}{6^2}+\frac{y^2}{2^2}=1 \end{aligned}$\begin{aligned} & \text{ length of } L R=\frac{2 b^2}{a}=\frac{2.2^2}{6}=\frac{8}{6}=\frac{4}{3}=\frac{m}{n} \\ & \therefore 2 m+n=2(4)+3=11\end{aligned}
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