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 $\frac{h}{a} + \frac{k}{b} = 1.$Also,the slope of the line $OP$ is $\frac{k}{h}$ and the slope of the given line is $-\frac{b}{a}.$Since $OP$ is perpendicular to the line,their product of slopes is $-1$: $\left(\frac{k}{h}\right) \left(-\frac{b}{a}\right) = -1$ $\Rightarrow \frac{b}{a} = \frac{h}{k}$ $\Rightarrow a = \frac{bh}{k}.$Substituting $a$ into the line equation: $\frac{hk}{bh} + \frac{y}{b} = 1$ $\Rightarrow \frac{k}{b} + \frac{y}{b} = 1$ $\Rightarrow b = k + \frac{ky}{h}$ (This is not correct,let's use the standard approach).Alternative approach: The line is $\frac{x}{a} + \frac{y}{b} = 1.$ The foot of the perpendicular $(h, k)$ satisfies $h = \frac{a b^2}{a^2 + b^2}$ and $k = \frac{a^2 b}{a^2 + b^2}.$Then $h^2 + k^2 = \frac{a^2 b^2 (b^2 + a^2)}{(a^2 + b^2)^2} = \frac{a^2 b^2}{a^2 + b^2} = \frac{1}{\frac{1}{b^2} + \frac{1}{a^2}}.$Given $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{2c^2},$ we have $h^2 + k^2 = \frac{1}{1/(2c^2)} = 2c^2.$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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