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$\frac{x}{a}\,\, + \,\,\frac{y}{b}\,\, = \,\,1$ એ ચલિત રેખા છે કે જેથી $\frac{1}{{{a^2}}}\, + \,\,\frac{1}{{{b^2}}}\,\, = \,\,\frac{1}{{{c^2}}}$ તો ઉગમબિંદુમાંથી રેખા પરના લંબપાદનું બિંદુપથ :
$x^2 + y^2 - ax - by = 0$
$x^2 + y^2 + ax + by = a^2 + b^2$
$x^2 + y^2 = c^2$
$x^2 - y^2 = 2c^2$
Solution
The equation of the line is $\frac{x}{a}+\frac{y}{b}=1 \quad \ldots(1)$ where $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2} \quad \cdots(2)$
Here $a, b$ are parameter while $c$ is a constant.
Any line $\perp$ to (1) is $\frac{x}{a}-\frac{y}{b}+k=0$
If it passes through the origin then $k =0$.
$\therefore$ Equation of the line through the origin and $\perp$ to (1) is $\frac{ x }{ b }-\frac{ y }{ a }=0 \quad \ldots(3)$
The locus of the foot of the $\perp$ from origin on $(1)$
i.e. locus of the point on intersection of $(1)$ and $(3)$ is obtained by eliminating
the parameters $a$ and $b$ between them.
Squaring $(1)$ and $(3)$ and adding, we get
$\left(\frac{1}{a^2}+\frac{1}{b^2}\right) x^2+\left(\frac{1}{b^2}+\frac{1}{a^2}\right) y^2=1 \Rightarrow \frac{x^2}{c^2}+\frac{y^2}{c^2}=1$ (Using (2))
$\Rightarrow x ^2+ y ^2= c ^2$, which is clearly a circle with center at origin and radius $c$.
