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  • Let P be a point on the ellipse <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,0<b<a.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7

Let P be a point on the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,0<b<a.} Let the line parallel to y-axis, passing through P meets the circle \mathrm{x^2+y^2=a^2} at the point Q such that P and Q are on the same side of x-axis. For two positive real numbers r and s, find the locus of the point R on PQ such that \mathrm{P R: R Q=r: s} and P varies over the

Option: 1

\mathrm{\frac{x^2}{a^2}+\frac{(r+s)^2 y^2}{(r a+s b)^2}=1}


Option: 2

\mathrm{\frac{x^2}{a^2}+\frac{(r-s)^2 y^2}{(r a+s b)^2}=1}


Option: 3

\mathrm{\frac{x^2}{a^2}+\frac{(r+s) y^2}{(b r+a s)^2}=1}


Option: 4

\mathrm{\frac{x^2}{a^2}+\frac{(r+s) y^2}{(a+b)^2}=1}


Answers (1)

best_answer

Let the coordinates of P be \mathrm{(a \cos \theta, b \sin \theta)}. Then the coordinates of Q are \mathrm{(a \cos \theta, a \sin \theta)}. \mathrm{R(h, k)} divides PQ in the ratio \mathrm{r: s}
\mathrm{ \begin{aligned} & \text { Then } R=\left(\frac{r a \cos \theta+s a \cos \theta}{r+s}, \frac{r a \sin \theta+s b \sin \theta}{r+s}\right) \\\\ & =\left(a \cos \theta, \frac{(r a+s b) \sin \theta}{r+s}\right)=(h, k) \\\\ & \therefore \cos \theta=\frac{h}{a}, \sin \theta=\frac{k(r+s)}{r a+s b} \end{aligned} }

Squaring and adding, \mathrm{\frac{h^2}{a^2}+\frac{(r+s)^2 k^2}{(r a+s b)^2}=1}

Hence locus of R is \mathrm{\frac{x^2}{a^2}+\frac{(r+s)^2 y^2}{(r a+s b)^2}=1.}

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avinash.dongre

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