JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the maximum area of the triangle that can be inscribed in the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1\), a \(>2\), having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the \(y\)-axis, be \(6 \sqrt{3}\). Then the eccentricity of the ellispe is
- A \(\frac{\sqrt{3}}{2}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{\sqrt{2}}\)
- D \(\frac{\sqrt{3}}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{3}}{2}\)
Step-by-step Solution
Detailed explanation
\(A=\frac{1}{2} a(1-\cos \theta)(4 \sin \theta)\) \(A =2 a (1-\cos \theta) \sin \theta\) \(\frac{ dA }{ d \theta}=2 a \left(\sin ^{2} \theta+\cos \theta-\cos ^{2} \theta\right)\) \(\frac{ dA }{ d \theta}=0 \Rightarrow 1+\cos \theta-2 \cos ^{2} \theta=0\)…
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