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If for a variable line $\frac{x}{a}+\frac{y}{b}=1$, the condition $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ (c is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is
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The correct answer is:
$x^2+y^2=c^2$
$\begin{array}{ll}
\Rightarrow b y-a x=0 \Rightarrow \frac{x}{b}-\frac{y}{a}=0 & {\left[\boxtimes m \text { of given line }=\frac{-b}{a}, \therefore m \text { of perpendicular }=\frac{a}{b}\right]}
\end{array}$
Now, the locus of foot of perpendicular is the intersection point of line $\frac{x}{a}+\frac{y}{b}=1$
and
$\frac{x}{b}-\frac{y}{a}=0$
To find locus, squaring and adding (i) and (ii)
$\begin{gathered}
\left(\frac{x}{a}+\frac{y}{b}\right)^2+\left(\frac{x}{b}-\frac{y}{a}\right)^2=1 \\
\Rightarrow \quad x^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+y^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)=1
\end{gathered}$
$\begin{aligned} & x^2\left(\frac{1}{c^2}\right)+y^2\left(\frac{1}{c^2}\right)=1,\left[0 \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}\right] \\ \Rightarrow & x^2+y^2=c^2\end{aligned}$
\Rightarrow b y-a x=0 \Rightarrow \frac{x}{b}-\frac{y}{a}=0 & {\left[\boxtimes m \text { of given line }=\frac{-b}{a}, \therefore m \text { of perpendicular }=\frac{a}{b}\right]}
\end{array}$
Now, the locus of foot of perpendicular is the intersection point of line $\frac{x}{a}+\frac{y}{b}=1$
and
$\frac{x}{b}-\frac{y}{a}=0$
To find locus, squaring and adding (i) and (ii)
$\begin{gathered}
\left(\frac{x}{a}+\frac{y}{b}\right)^2+\left(\frac{x}{b}-\frac{y}{a}\right)^2=1 \\
\Rightarrow \quad x^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+y^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)=1
\end{gathered}$
$\begin{aligned} & x^2\left(\frac{1}{c^2}\right)+y^2\left(\frac{1}{c^2}\right)=1,\left[0 \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}\right] \\ \Rightarrow & x^2+y^2=c^2\end{aligned}$
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