Find the maximum area of an isosceles triangl | vedclass

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(A) The given ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Let the vertex $C$ be at $(a, 0)$. Let the other two vertices be $A(-x_1, y_1)$ and $

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$\frac{3\sqrt{3}}{4} ab$

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(A) The given ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Let the vertex $C$ be at $(a, 0)$. Let the other two vertices be $A(-x_1, y_1)$ and $B(-x_1, -y_1)$ where $x_1 > 0$ and $y_1 > 0$.Since $A(-x_1, y_1)$ lies on the ellipse,we have $y_1 = \frac{b}{a} \sqrt{a^2 - x_1^2}$.The base of the triangle is $2y_1 = \frac{2b}{a} \sqrt{a^2 - x_1^2}$ and the height is $h = a - (-x_1) = a + x_1$.The area $A$ of the triangle is $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \left( \frac{2b}{a} \sqrt{a^2 - x_1^2} ight) 	imes (a + x_1) = \frac{b}{a} (a + x_1) \sqrt{(a - x_1)(a + x_1)} = \frac{b}{a} (a + x_1)^{3/2} (a - x_1)^{1/2}$.To maximize $A$,we maximize $A^2 = \frac{b^2}{a^2} (a + x_1)^3 (a - x_1)$.Let $f(x_1) = (a + x_1)^3 (a - x_1)$. Then $f'(x_1) = 3(a + x_1)^2 (a - x_1) - (a + x_1)^3 = (a + x_1)^2 [3a - 3x_1 - a - x_1] = (a + x_1)^2 (2a - 4x_1)$.Setting $f'(x_1) = 0$,we get $x_1 = a/2$ (since $x_1 = -a$ is not possible).Substituting $x_1 = a/2$ into the area formula:$A = \frac{b}{a} (a + a/2) \sqrt{a^2 - a^2/4} = \frac{b}{a} (\frac{3a}{2}) \sqrt{\frac{3a^2}{4}} = \frac{b}{a} \cdot \frac{3a}{2} \cdot \frac{a\sqrt{3}}{2} = \frac{3\sqrt{3}}{4} ab$.

Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with its vertex at one end of the major axis.

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