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  • The total area bounded by the locus of mid point of PS and the ellipse <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5

The total area bounded by the locus of mid point of PS and the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}, where P be any point on the ellipse and S be the focus of the ellipse, is (area measured in square units)

Option: 1

\mathrm{\frac{25}{16} \pi^2 a^2 b^2}


Option: 2

\mathrm{\frac{5}{4} \pi a b}


Option: 3

\mathrm{\frac{3 \pi}{4} a b}


Option: 4

\mathrm{\frac{4 \pi}{5} a b}


Answers (1)

best_answer

Given, equation of ellipse is \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}

\mathrm{\therefore \quad}  Parametric coordinates of \mathrm{P=(a \cos \theta, b \sin \theta)}

Focus of ellipse \mathrm{=S(a e, 0)}

Also, area of ellipse \mathrm{=\pi a b} sq. units

Now, let \mathrm{M(h, k)} be mid point of PS.

\mathrm{ \begin{aligned} & \therefore \quad h=\frac{a \cos \theta+a e}{2}, k=\frac{b \sin \theta}{2} \\\\ & \therefore \quad \cos \theta=\frac{\left(h-\frac{a e}{2}\right)}{a / 2} \text { and } \sin \theta=\frac{2 k}{b} \text { or } \frac{k}{b / 2} \\\\ & \because \quad \sin ^2 \theta+\cos ^2 \theta=1, \text { therefore } \\\\ & \frac{k^2}{b^2 / 4}+\frac{\left(h-\frac{a e}{2}\right)^2}{a^2 / 4}=1 \text { or } \frac{y^2}{b^2 / 4}+\frac{\left(x-\frac{a e}{2}\right)^2}{a^2 / 4}=1 \end{aligned} }
which is an ellipse or locus of \mathrm{M(h, k)} whose area is

\mathrm{=\pi\left(\frac{a}{2} \times \frac{b}{2}\right)=\frac{\pi a b}{4}} sq. units

\mathrm{ \therefore \quad \text { Total area }=\pi a b\left(1+\frac{1}{4}\right)=\frac{5 \pi a b}{4} }

Posted by

sudhir kumar

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