frac{x}{a},, + ,,frac{y}{b},, = ,,1 એ ચલિત રે | vedclass

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Question

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,

Vedclass · Accepted Answer

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

Vedclass · Answer

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 $k =0$.

$	herefore$ 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

$\left(\frac{1}{a^2}+\frac{1}{b^2}ight) x^2+\left(\frac{1}{b^2}+\frac{1}{a^2}ight) y^2=1 \Rightarrow \frac{x^2}{c^2}+\frac{y^2}{c^2}=1$ (Using (2))

$\Rightarrow x ^2+ y ^2= c ^2$, which is clearly a circle with center at origin and radius $c$.

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