JEE Advance - Mathematics (2001 - No. 12)
Let $$P$$ be a point on the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,0 < b < a$$. Let the line parallel to $$y$$-axis passing through $$P$$ meet the circle $${x^2} + {y^2} = {a^2}$$ 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 $$PR$$ : $$RQ = r: s$$ as $$P$$ varies over the ellipse.
${{{x^2}} \over {{a^2}}} + {{{y^2}{{\left( {r + s} \right)}^2}} \over {{{\left( {bs + ar} \right)}^2}}} = 1$
${{{x^2}} \over {{a^2}}} - {{{y^2}{{\left( {r + s} \right)}^2}} \over {{{\left( {bs + ar} \right)}^2}}} = 1$
${{{x^2}} \over {{a^2}}} + {{{y^2}{{\left( {r - s} \right)}^2}} \over {{{\left( {bs + ar} \right)}^2}}} = 1$
${{{x^2}} \over {{a^2}}} + {{{y^2}{{\left( {r + s} \right)}^2}} \over {{{\left( {bs - ar} \right)}^2}}} = 1$
${{{x^2}} \over {{a^2}}} + {{{y^2}{{\left( {r + s} \right)}^2}} \over {{{\left( {as + br} \right)}^2}}} = 1$
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