The locus of the foot of the perpendicular dr | vedclass

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(D) Let the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos 	heta, b \sin 	heta)$ be $\frac{x \cos 	heta}{a} +

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos 	heta, b \sin 	heta)$ be $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$.Let $(h, k)$ be the foot of the perpendicular from the centre $(0, 0)$ to this tangent.The slope of the tangent is $m_1 = -\frac{b \cos 	heta}{a \sin 	heta}$.The slope of the line joining $(0, 0)$ and $(h, k)$ is $m_2 = \frac{k}{h}$.Since the lines are perpendicular,$m_1 	imes m_2 = -1$,so $(-\frac{b \cos 	heta}{a \sin 	heta}) 	imes (\frac{k}{h}) = -1$,which gives $\frac{k}{h} = \frac{a \sin 	heta}{b \cos 	heta}$.This implies $\frac{h}{a \cos 	heta} = \frac{k}{b \sin 	heta} = \lambda$.Substituting into the tangent equation: $\lambda \cos^2 	heta + \lambda \sin^2 	heta = 1$,so $\lambda = 1$.Thus,$h = a \cos 	heta$ and $k = b \sin 	heta$,which means $\cos 	heta = \frac{h}{a}$ and $\sin 	heta = \frac{k}{b}$.Using $\cos^2 	heta + \sin^2 	heta = 1$,we get $\frac{h^2}{a^2} + \frac{k^2}{b^2} = 1$.However,the foot of the perpendicular $(h, k)$ also lies on the tangent,so $\frac{h^2}{a} + \frac{k^2}{b} = h \cos 	heta + k \sin 	heta$ is not the path. The correct relation is derived from $h^2 + k^2 = (a \cos 	heta)^2 + (b \sin 	heta)^2$ is not correct. The correct locus is $(x^2 + y^2)^2 = a^2x^2 + b^2y^2$.

The 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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