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Application of Derivatives
NCERT Class 12
CBSE
Grade 12
Question:
Find the maximum area of an isosceles triangle inscribed in the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with its vertex at one end of the major axis.
Step-by-Step Solution
$A = ab \sin\theta(1 - \cos\theta) \Rightarrow \cos\theta = -1/2$. [1.5 Marks] Max Area $= ab(\sqrt{3}/2)(3/2) = \dfrac{3\sqrt{3}}{4} ab$. [1.5 Marks]
--- 🎯 Official CBSE Marking Scheme: Setting up area function in terms of parametric angle $\theta$: 1.5 Marks Evaluating maximum area $= \frac{3\sqrt{3}}{4} ab$: 1.5 Marks
Correct Answer:
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