Question

If a perpendicular be let fall from any point P upon its polar prove that the distance of the foot of this perpendicular from the focus is equal to the distance of the point P from the directrix.

Answer 1

Consider the equation of the parabola to be $y^2=4ax$ ......(1)

The focus of it will be (a,0) and directrix will be $x+a=0$ ......(2)

If P is considered to be any point $(x_1,y_1)$, the equation of its polar to (1) will be,

$y{y}_1=2a(x+x_1)$ ......(3)

Equation to the line passing through P and perpendicular to (3) will be,

$y-y_1=-\frac{y_1}{2a}(x-x_1)=2ay=-y_{1}x+x_1{y}_1+2a{y}_1$ ......(4)

Solve the above equation to get the result of x and y as,

$x=-\frac{x_1{y}_2^2+2a{y}_1^2-4a^2{x}_1}{y_1^2+4a^2},y=\frac{4a{y}_1(x_2+a)}{y_1^2+4a^2}$

So, the coordinates of the point of intersection of (3) and (4) which is the foot of the perpendicular to polar Q are

$[\frac{x_1{y}_2^2+2a{y}_1^2-4a^2{x}_1}{y_1^2+4a^2},\frac{4ay{}_1(x_1+a)}{y_1^2+4a^2}]$

So, the distance of the focus and Q will be,

$\sqrt{[{\left\{\frac{x_1{y}_2^2+a{y}_1^2-4a^2{x}_1-4a^3}{y_1^2+4a^2}\right\}}^2+{\left\{\frac{4ay{}_1(x_1+a)}{y_1^2+4a^2}\right\}}^2]}=\frac{1}{y_1^2+4a^2}\sqrt{[{(x_1{y}_1^2+a{y}_1^2-4a^2{x}_1-4a^3)}^2+16a^2{y}_1^2{(x_1+a)}^2]}=\frac{1}{y_1^2+4a^2}\sqrt{[{(y_1^2+4a^2)}^2{(x_1+a)}^2+16a^2{y}_1^2{(x_1+a)}^2]}=\frac{(x_1+a)}{y_1^2+4a^2}\sqrt{{(y_1^2-4a^2)}^2}+16a^2+y_1^2=\frac{(x_1+a)}{y_1^2+4a^2}\times (y_1^2+4a^2)=x_1+a$

Hence, the distance of the point $P(x_1,y_1)$, from the directrix $x+a=0$ is clearly $(x_1+a)$ which is equal to the distance that is given by (5).