Centre of the circle =(23,1) Equation of diameter =2x+3y−k=0 $\begin{aligned}
& 2\left(\frac{3}{2}\right)+3(1)-\mathrm{k}=0 \
& \Rightarrow \mathrm{k}=6
\end{aligned}Now,Equationofellipsebecomes\begin{aligned}
& x^2+9 y^2=36 \
& \frac{x^2}{6^2}+\frac{y^2}{2^2}=1
\end{aligned}\begin{aligned} & \text { length of } L R=\frac{2 b^2}{a}=\frac{2.2^2}{6}=\frac{8}{6}=\frac{4}{3}=\frac{m}{n} \ & \therefore 2 m+n=2(4)+3=11\end{aligned}$
Let the line 2 x+3 y- k=0, k>0, intersect the x-axis and y-axis at the points A and B, respectively. If the equation of the circle having the line…
JEE Main 2024 — Mathematics Coordinate Geometry
Let the line 2x+3y−k=0,k>0, intersect the x-axis and y-axis at the points A and B, respectively. If the equation of the circle having the line segment AB as a diameter is x2+y2−3x−2y=0 and the length of the latus rectum of the ellipse x2+9y2=k2 is nm, where m and n are coprime, then 2 m+n is equal to
Held on 5 Apr 2024 · Verified 6 Jul 2026.
Options
- A
11
- B
10
- C
12
- D
13
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