Locus of the foot of the perpendicular drawn | vedclass

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(d) $\frac{{x\cos 	heta }}{a} + \frac{{y\sin 	heta }}{b} = 1,$ Slope $ \equiv - \frac{{b\cot 	heta }}{a} 	imes \frac{k}{h} = - 1$

$\frac{{h\cos

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) $\frac{{x\cos 	heta }}{a} + \frac{{y\sin 	heta }}{b} = 1,$ Slope $ \equiv - \frac{{b\cot 	heta }}{a} 	imes \frac{k}{h} = - 1$

$\frac{{h\cos 	heta }}{a} + \frac{{k\sin 	heta }}{b} = 1$

Since ${m{cosec}}	heta = \sqrt {1 + \frac{{{a^2}{h^2}}}{{{k^2}{b^2}}}} $

==> $\frac{{{h^2}}}{{kb}} + \frac{k}{b} = {m{cosec}}	heta $

==> $({h^2} + {k^2}) = bk\,{m{cosec}}	heta $

= $\frac{{bk(\sqrt {{k^2}{b^2} + {a^2}{h^2}} )}}{{bk}}$

==> ${({x^2} + {y^2})^2} = {a^2}{x^2} + {b^2}{y^2}$.

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

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