P is a point on the conic a^2 x^2+b^2 y^2=a^2 | vedclass

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(B) Given the equation of the conic:$a^2 x^2 + b^2 y^2 = a^2(a^2 + b^2 - y^2)$$\Rightarrow a^2 x^2 + b^2 y^2 + a^2 y^2 = a^2(a^2 + b^2)$$\Rightarrow a

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$\frac{\sqrt{a^2+b^2}}{b}$

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(B) Given the equation of the conic:$a^2 x^2 + b^2 y^2 = a^2(a^2 + b^2 - y^2)$$\Rightarrow a^2 x^2 + b^2 y^2 + a^2 y^2 = a^2(a^2 + b^2)$$\Rightarrow a^2 x^2 + y^2(a^2 + b^2) = a^2(a^2 + b^2)$Dividing by $a^2(a^2 + b^2)$:$\frac{x^2}{a^2 + b^2} + \frac{y^2}{a^2} = 1$This is an ellipse with $A^2 = a^2 + b^2$ and $B^2 = a^2$. Since $A^2 > B^2$,the eccentricity $e$ is given by:$e = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{a^2}{a^2 + b^2}} = \sqrt{\frac{b^2}{a^2 + b^2}} = \frac{b}{\sqrt{a^2 + b^2}}$By the definition of a conic section,the distance from a point $P$ to the focus $S$ is $SP = e \cdot PM$,where $PM$ is the perpendicular distance to the directrix.Thus,$PM = \frac{1}{e} SP$.Comparing this with $PM = K SP$,we get $K = \frac{1}{e}$.$K = \frac{\sqrt{a^2 + b^2}}{b}$.

$P$ is a point on the conic $a^2 x^2+b^2 y^2=a^2(a^2+b^2-y^2)$ and $S$ is a focus of that conic. $M$ is the foot of the perpendicular from $P$ onto a directrix of that conic nearer to $S$. If $PM = K SP$,then $K=$

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