Find the area bounded by the ellipse frac{x^{ | vedclass

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(B) The area of the region bounded by the ellipse and the ordinates $x=0$ and $x=ae$ is symmetric about the $x$-axis.
The area is given by $A = 2 \in

Vedclass · Accepted Answer

$ab[e\sqrt{1-e^{2}}+\sin^{-1}e]$

Vedclass · Answer

(B) The area of the region bounded by the ellipse and the ordinates $x=0$ and $x=ae$ is symmetric about the $x$-axis.
The area is given by $A = 2 \int_{0}^{ae} y \, dx$.
From the equation of the ellipse,$y = \frac{b}{a} \sqrt{a^{2}-x^{2}}$.
Substituting this into the integral,we get $A = 2 \frac{b}{a} \int_{0}^{ae} \sqrt{a^{2}-x^{2}} \, dx$.
Using the standard integral $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{x}{a})$,we have:
$A = \frac{2b}{a} \left[ \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{x}{a}) \right]_{0}^{ae}$
$A = \frac{2b}{a} \left[ \frac{ae}{2} \sqrt{a^{2}-a^{2}e^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{ae}{a}) - (0 + 0) \right]$
$A = \frac{2b}{a} \left[ \frac{ae}{2} \cdot a\sqrt{1-e^{2}} + \frac{a^{2}}{2} \sin^{-1}(e) \right]$
$A = \frac{2b}{a} \cdot \frac{a^{2}}{2} \left[ e\sqrt{1-e^{2}} + \sin^{-1}(e) \right]$
$A = ab[e\sqrt{1-e^{2}} + \sin^{-1}(e)]$.

Find the area bounded by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the ordinates $x=0$ and $x=ae$,where $b^{2}=a^{2}(1-e^{2})$ and $e < 1$.

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