The line frac{x}{a} + frac{y}{b} = 1 moves in | vedclass

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(B) Let the foot of the perpendicular from the origin $(0,0)$ to the line $\frac{x}{a} + \frac{y}{b} = 1$ be $P(h, k)$.Since $P(h, k)$ lies on the lin

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$x^2 + y^2 = 2c^2$

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(B) Let the foot of the perpendicular from the origin $(0,0)$ to the line $\frac{x}{a} + \frac{y}{b} = 1$ be $P(h, k)$.Since $P(h, k)$ lies on the line,we have $\frac{h}{a} + \frac{k}{b} = 1$.The line $OP$ is perpendicular to the given line. The slope of the given line is $-\frac{b}{a}$,and the slope of $OP$ is $\frac{k}{h}$.Thus,$(-\frac{b}{a}) 	imes (\frac{k}{h}) = -1$ $\Rightarrow \frac{b}{a} = \frac{h}{k}$ $\Rightarrow a = \frac{bh}{k}$.Substituting $a$ into the line equation: $\frac{hk}{bh} + \frac{k}{b} = 1$ $\Rightarrow \frac{k}{b} + \frac{k}{b} = 1$ $\Rightarrow \frac{2k}{b} = 1$ $\Rightarrow b = 2k$. Similarly,$a = 2h$.Given $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2}$,substituting $a=2h$ and $b=2k$ gives $\frac{1}{4h^2} + \frac{1}{4k^2} = \frac{1}{2c^2}$.$\frac{h^2 + k^2}{4h^2k^2} = \frac{1}{2c^2} \Rightarrow h^2 + k^2 = \frac{2h^2k^2}{c^2}$.Wait,the standard condition for the foot of the perpendicular $(h,k)$ is $h = a \cos^2 	heta, k = b \sin^2 	heta$ where $\frac{x}{a} + \frac{y}{b} = 1$ is $x \cos 	heta + y \sin 	heta = p$. Here $p = \frac{1}{\sqrt{1/a^2 + 1/b^2}}$.Since $P(h,k)$ is the foot of the perpendicular,$h^2+k^2 = p^2 = \frac{1}{1/a^2 + 1/b^2} = \frac{1}{1/(2c^2)} = 2c^2$.Thus,the locus is $x^2 + y^2 = 2c^2$.

The line $\frac{x}{a} + \frac{y}{b} = 1$ moves in such a way that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2}$,where $a, b, c \in R_0$ and $c$ is a constant. Then,the locus of the foot of the perpendicular from the origin on the given line is -

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