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(A) Let the equation of the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be $y = mx + \sqrt{a^2 m^2 + b^2}$.Let the foot of the perpendi

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$(x^2+y^2)^2=a^2 x^2+b^2 y^2$

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(A) Let the equation of the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be $y = mx + \sqrt{a^2 m^2 + b^2}$.Let the foot of the perpendicular from the centre $(0, 0)$ to this tangent be $(h, k)$.The slope of the tangent is $m$,so the slope of the perpendicular line is $-\frac{1}{m}$.The equation of the perpendicular line passing through $(0, 0)$ is $y = -\frac{1}{m} x$,which implies $m = -\frac{x}{y}$.Since $(h, k)$ lies on the tangent,$k = mh + \sqrt{a^2 m^2 + b^2}$.Substituting $m = -\frac{h}{k}$ into the equation: $k = -\frac{h^2}{k} + \sqrt{a^2 \frac{h^2}{k^2} + b^2}$.$k + \frac{h^2}{k} = \sqrt{\frac{a^2 h^2 + b^2 k^2}{k^2}}$.$\frac{k^2 + h^2}{k} = \frac{\sqrt{a^2 h^2 + b^2 k^2}}{k}$.Squaring both sides,we get $(h^2 + k^2)^2 = a^2 h^2 + b^2 k^2$.Replacing $(h, k)$ with $(x, y)$,the locus is $(x^2 + y^2)^2 = a^2 x^2 + b^2 y^2$.

The equation of the locus of the foot of the perpendicular drawn from the centre of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ to any tangent of the ellipse is

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