JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the line \(2 \mathrm{x}+3 \mathrm{y}-\mathrm{k}=0, \mathrm{k}>0\), intersect the \(\mathrm{x}\)-axis and \(\mathrm{y}\)-axis at the points \(\mathrm{A}\) and \(\mathrm{B}\), respectively. If the equation of the circle having the line segment \(\mathrm{AB}\) as a diameter is \(\mathrm{x}^2+\mathrm{y}^2-3 \mathrm{x}-2 \mathrm{y}=0\) and the length of the latus rectum of the ellipse \(\mathrm{x}^2+9 \mathrm{y}^2=\mathrm{k}^2\) is \(\frac{\mathrm{m}}{\mathrm{n}}\), where \(\mathrm{m}\) and \(\mathrm{n}\) are coprime, then \(2 \mathrm{~m}+\mathrm{n}\) is equal to
- A \(10\)
- B \(11\)
- C \(13\)
- D \(12\)
Answer & Solution
Correct Answer
(B) \(11\)
Step-by-step Solution
Detailed explanation
Centre of the circle \(=\left(\frac{3}{2}, 1\right) \) Equation of diameter \(=2 \mathrm{x}+3 \mathrm{y}-\mathrm{k}=0 \) \( 2\left(\frac{3}{2}\right)+3(1)-\mathrm{k}=0 \) \( \Rightarrow \mathrm{k}=6\) Now, Equation of ellipse becomes \( x^2+9 y^2=36 \)…
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