Question

A variable line makes intercept on the cordinate axes the sum of whose squares is constant and is equal to $a^2$. Find the locus of the foot of the perpendicular from the origin to this line.

Answer 1

The equation of line is $\frac{x}{p}+\frac{y}{q}=1$
Here $p\&q$ are the intercepts an the axis
$p^2+q^2=a^2$
Let foot of perpendicular be $x_1,y_1$
Which must line on $\frac{x}{p}+\frac{y}{q}=1$
We also can include that.
$\frac{y_1}{x_1}\times \frac{-q}{p}=-1p^2=a^2-q^2\frac{y_1}{x_1}=\frac{\sqrt{a^2-q^2}}{q}\frac{q}{\sqrt{a^2-q^2}}=\frac{x_1}{y_1}q=\frac{x_1\sqrt{a^2-q^2}}{y_1}p=a^2-\frac{x_1\sqrt{a^2-q^2}}{y_1}p=\frac{a^2{y}_1-x_1\sqrt{a^2-q^2}}{y_1}\frac{x}{p}+\frac{y}{q}=1\frac{x_1{y}_1}{a^2-x_1\sqrt{a^2-q^2}}+\frac{x_1{y}_1}{x_1\sqrt{a^2-q^2}}=1\frac{{x_1}^2{y}_1\sqrt{a^2-q^2}+a^2{x}_1{y}_1-{x_1}^2{y}_1\sqrt{a^2-q^2}}{(a^2-x_1\sqrt{a^2-q^2})\left(x_1\sqrt{a^2-q^2}\right)}=1a^2{x}_1{y}_1=a^2{x}_1\sqrt{a^2-q^2}-{x_1}^2{a}^2+{x_1}^2{q}^2{x_1}^2{a}^2+{x_1}^2{q}^2+a^2{x}_1{y}_1-a^2{x}_1\sqrt{a^2-q^2}=0x_1(a^2{x}_1+q^2{x}_1+a^2{y}_1-a^2\sqrt{a^2-q^2})=0{x_{1}a}^2+{x_{1}q}^2+y_1{a}^2-a^{2}p=0a^2(x_1+y_1)+{x_{1}q}^2-a^{2}p=0$