Let P be a point on the ellipse frac{x^2}{9}+ | vedclass

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(C) Let the point $P$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $(a \cos 	heta, b \sin 	heta)$. Here $a^2=9$ and $b^2=4$,so $a=3, b=

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$x^2+y^2=25$

Vedclass · Answer

(C) Let the point $P$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $(a \cos 	heta, b \sin 	heta)$. Here $a^2=9$ and $b^2=4$,so $a=3, b=2$. Thus $P = (3 \cos 	heta, 2 \sin 	heta)$.The auxiliary circle is $x^2 + y^2 = a^2 = 9$. The point $Q$ on the auxiliary circle corresponding to $P$ is $(3 \cos 	heta, 3 \sin 	heta)$.The normal to the ellipse at $P(3 \cos 	heta, 2 \sin 	heta)$ is $\frac{ax}{\cos 	heta} - \frac{by}{\sin 	heta} = a^2 - b^2$,which is $\frac{3x}{\cos 	heta} - \frac{2y}{\sin 	heta} = 9 - 4 = 5$.The normal to the circle at $Q(3 \cos 	heta, 3 \sin 	heta)$ passes through the center $(0,0)$ and $Q$,so its equation is $y = x 	an 	heta$,or $x \sin 	heta - y \cos 	heta = 0$.Let $R = (h, k)$. Since $R$ lies on both normals,we have $\frac{3h}{\cos 	heta} - \frac{2k}{\sin 	heta} = 5$ and $h \sin 	heta = k \cos 	heta$.From $h \sin 	heta = k \cos 	heta$,we get $\sin 	heta = \frac{k \cos 	heta}{h}$. Substituting into the first equation: $\frac{3h}{\cos 	heta} - \frac{2kh}{k \cos 	heta} = 5 \implies \frac{3h-2h}{\cos 	heta} = 5 \implies \cos 	heta = \frac{h}{5}$.Then $\sin 	heta = \frac{k}{h} \cdot \frac{h}{5} = \frac{k}{5}$.Using $\sin^2 	heta + \cos^2 	heta = 1$,we get $(\frac{k}{5})^2 + (\frac{h}{5})^2 = 1$,which simplifies to $h^2 + k^2 = 25$.Thus,the locus of $R$ is $x^2 + y^2 = 25$.

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