Question

If the line $\frac{x}{a}+\frac{y}{b}=1$ moves 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

• A. Straight line
• B. Circle
• C. Parabola
• D. Ellipse

Answer 1

The correct option is B Circle

Equation of line is $\frac{x}{a}+\frac{y}{b}=1....\left(i\right)$
Let the foot of the perpendicular drawn from the origin to the line be $P(x_1,y_1)$.
Since $OP\perp AB$
$\therefore$ slope of $OP\times$ slope of $AB=1$
$\Rightarrow \frac{y_1}{x_1}\times \frac{b}{-a}=-1\Rightarrow b{y}_1=a{x}_1....\left(ii\right)$
Since, $P$ lies on the line $AB$, so
$\frac{x_1}{a}+\frac{y_1}{b}=1$
$\Rightarrow b{x}_1+a{y}_1=ab....\left(iii\right)$
From (ii) and (iii), we get
$x_1=\frac{a{b}^2}{a^2+b^2}$ and $y_1=\frac{a^{2}b}{a^2+b^2}$
Now, $x_1^2+y_1^2={\left(\frac{a{b}^2}{a^2+b^2}\right)}^2+{\left(\frac{a^{2}b}{a^2+b^2}\right)}^2$
$=\frac{a^2{b}^4}{(a^2+b^2{)}^2}+\frac{a^4{b}^2}{(a^2+b^2{)}^2}=\frac{a^2{b}^2}{(a^2+b^2{)}^2}(a^2+b^2)$
$=\frac{a^2{b}^2}{a^2+b^2}=\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}}$
$\Rightarrow x_1^2+y_1^2=\frac{1}{C^2}(\because \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2})$
Thus, the locus of $P(x_1,y_1)$ is $x^2+y^2=c^2$ which is a circle.