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Locus of the foot of the perpendicular drawn from the centre upon any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, is
${({x^2} + {y^2})^2} = {b^2}{x^2} + {a^2}{y^2}$
${({x^2} + {y^2})^2} = {b^2}{x^2} - {a^2}{y^2}$
${({x^2} + {y^2})^2} = {a^2}{x^2} - {b^2}{y^2}$
${({x^2} + {y^2})^2} = {a^2}{x^2} + {b^2}{y^2}$
Solution
(d) $\frac{{x\cos \theta }}{a} + \frac{{y\sin \theta }}{b} = 1,$ Slope $ \equiv – \frac{{b\cot \theta }}{a} \times \frac{k}{h} = – 1$
$\frac{{h\cos \theta }}{a} + \frac{{k\sin \theta }}{b} = 1$
Since ${\rm{cosec}}\theta = \sqrt {1 + \frac{{{a^2}{h^2}}}{{{k^2}{b^2}}}} $
==> $\frac{{{h^2}}}{{kb}} + \frac{k}{b} = {\rm{cosec}}\theta $
==> $({h^2} + {k^2}) = bk\,{\rm{cosec}}\theta $
= $\frac{{bk(\sqrt {{k^2}{b^2} + {a^2}{h^2}} )}}{{bk}}$
==> ${({x^2} + {y^2})^2} = {a^2}{x^2} + {b^2}{y^2}$.
