Calculus

Application of Integrals

Hard

core

The area bounded by an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the lines $x=0, x=ae$ and the $x$ -axis in the first quadrant, where $e$ is the eccentricity of the ellipse, is:

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

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

$\frac{a}{b}\left[\sin^{-1}e+e\sqrt{1-e^2}\right]$

$\frac{ab}{2}\left[\sin^{-1}e-e\sqrt{1-e^2}\right]$

✅ Correct Option: 1

Area $=\int_0^{ae}\frac{b}{a}\sqrt{a^2-x^2}\,dx=\frac{b}{a}\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^{ae}=\frac{ab}{2}\left[\sin^{-1}e+e\sqrt{1-e^2}\right]$ .

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The area bounded by an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1a2x2​+b2y2​=1 and the lines x=0,x=aex=0, x=aex=0,x=ae and the xxx-axis in the first quadrant, where eee is the eccentricity of the ellipse, is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The area bounded by an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the lines $x=0, x=ae$ and the $x$ -axis in the first quadrant, where $e$ is the eccentricity of the ellipse, is: