If the line frac{x}{a} + frac{y}{b} = 1 is a | vedclass

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(C) The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1 \quad \dots(1)$ where $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \quad \dots(2)$.Le

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

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(C) The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1 \quad \dots(1)$ where $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \quad \dots(2)$.Let $(h, k)$ be the foot of the perpendicular from the origin $(0, 0)$ to the line $(1)$.The line passing through $(0, 0)$ and $(h, k)$ is perpendicular to the line $(1)$.The slope of line $(1)$ is $m_1 = -\frac{b}{a}$.The slope of the line passing through $(0, 0)$ and $(h, k)$ is $m_2 = \frac{k}{h}$.Since the lines are perpendicular,$m_1 	imes m_2 = -1 \implies (-\frac{b}{a}) 	imes (\frac{k}{h}) = -1 \implies \frac{b}{a} = \frac{h}{k} \implies \frac{a}{h} = \frac{b}{k} = \lambda$.Thus,$a = h\lambda$ and $b = k\lambda$.Since $(h, k)$ lies on line $(1)$,$\frac{h}{h\lambda} + \frac{k}{k\lambda} = 1 \implies \frac{1}{\lambda} + \frac{1}{\lambda} = 1 \implies \lambda = 2$.So,$a = 2h$ and $b = 2k$.Substituting these into equation $(2)$,we get $\frac{1}{(2h)^2} + \frac{1}{(2k)^2} = \frac{1}{c^2} \implies \frac{1}{4h^2} + \frac{1}{4k^2} = \frac{1}{c^2} \implies \frac{h^2 + k^2}{4h^2k^2} = \frac{1}{c^2}$.Wait,let's re-evaluate the perpendicular condition. The line is $\frac{x}{a} + \frac{y}{b} = 1$. The perpendicular line through origin is $bx - ay = 0$. The intersection $(h, k)$ satisfies $\frac{h}{a} + \frac{k}{b} = 1$ and $bh - ak = 0 \implies \frac{h}{a} = \frac{k}{b} = \frac{h^2+k^2}{ah+bk} = \frac{h^2+k^2}{a(h) + b(k)}$.Actually,using the property of the foot of the perpendicular $(h, k)$ for $\frac{x}{a} + \frac{y}{b} = 1$: $h = \frac{a b^2}{a^2+b^2}$ and $k = \frac{a^2 b}{a^2+b^2}$.Then $h^2 + k^2 = \frac{a^2 b^4 + a^4 b^2}{(a^2+b^2)^2} = \frac{a^2 b^2 (b^2 + a^2)}{(a^2+b^2)^2} = \frac{a^2 b^2}{a^2+b^2}$.Since $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \implies \frac{a^2+b^2}{a^2b^2} = \frac{1}{c^2} \implies \frac{a^2b^2}{a^2+b^2} = c^2$.Therefore,$h^2 + k^2 = c^2$. The locus is $x^2 + y^2 = c^2$.

If the line $\frac{x}{a} + \frac{y}{b} = 1$ is a variable line such that $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$,then the locus of the foot of the perpendicular from the origin to the line is:

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