Question

If the perpendicular is let fall from the point P on parabola upon its polar. Prove that the distance of the foot of this perpendicular from the focus is equal to the distance of point P from its directrix.

Answer 1

Hint: The rough figure drawn based on the given information is

We assume that equation of parabola as $$y^2=4ax$$ then equation of directrix is given as
$$x+a=0$$ then we assume that point P as $$(h,k)$$ to find the polar of point P with respect to parabola $$S\equiv y^2=4ax$$ given as $$S_1=0$$ where
$$S_1=yk-2a(x+h)$$
Then we find the foot of perpendicular of point $$P(h,k)$$ to its polar $$ax+by+c=0$$ given as
$${\displaystyle \frac{x-h}{a}}={\displaystyle \frac{y-k}{b}}={\displaystyle \frac{-(ah+bk+c)}{a^2+b^2}}$$
Then we find the distance between focus ‘S’ and foot of perpendicular ‘T’ using distance formula that is the formula of distance between $$(x_1,y_1),(x_2,y_2)$$ is given as
$$ST=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$$
Then we find the distance of point P to the directrix using perpendicular formula that is the distance of $$P(h,k)$$ to line $$ax+by+c=0$$ is given as
$$D={\displaystyle \frac{|ah+bk+c|}{\sqrt{a^2+b^2}}}$$

Complete step-by-step solution:
Let us assume that the equation of parabola as
$$\Rightarrow y^2=4ax$$
Let us assume the point P as $$P(h,k)$$
We know that the equation of polar of point P with respect to parabola $$S\equiv y^2=4ax$$ given as $$S_1=0$$ where
$$S_1=yk-2a(x+h)$$
By using the above formula we get the equation of polar of P as
$$\begin{array}{rl}& \Rightarrow yk-2a(x+h)=0\\ & \Rightarrow 2ax-ky+2ah=0\end{array}$$
Now, let us find the foot of perpendicular of $$P(h,k)$$ on its polar which is $${}^{\prime }{T}^{\prime }$$
We know that we can find the foot of perpendicular of point $$P(h,k)$$ to its polar $$ax+by+c=0$$ given as
$${\displaystyle \frac{x-h}{a}}={\displaystyle \frac{y-k}{b}}={\displaystyle \frac{-(ah+bk+c)}{a^2+b^2}}$$
By using the above formula we get the co – ordinates of point $${}^{\prime }{T}^{\prime }$$ as
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{x-h}{2a}}={\displaystyle \frac{y-k}{-k}}={\displaystyle \frac{-(2ah-k^2+2ah)}{4a^2+k^2}}\\ & \Rightarrow {\displaystyle \frac{x-h}{2a}}={\displaystyle \frac{y-k}{-k}}={\displaystyle \frac{k^2-4ah}{4a^2+k^2}}\end{array}$$
By dividing the terms we get the $${}^{\prime }{x}^{\prime }$$ co – ordinate of $${}^{\prime }{T}^{\prime }$$ as
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{x-h}{2a}}={\displaystyle \frac{k^2-4ah}{4a^2+k^2}}\\ & \Rightarrow x=h+{\displaystyle \frac{2a{k}^2-8a^{2}h}{4a^2+k^2}}\end{array}$$
By doing the LCM and adding the terms we get
$$\begin{array}{rl}& \Rightarrow x={\displaystyle \frac{4a^{2}h+h{k}^2+2a{k}^2-8a^{2}h}{4a^2+k^2}}\\ & \Rightarrow x={\displaystyle \frac{h{k}^2+2a{k}^2-4a^{2}h}{4a^2+k^2}}\end{array}$$

Similarly, we get the $${}^{\prime }{y}^{\prime }$$ co – ordinate of $${}^{\prime }{T}^{\prime }$$ as
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{y-k}{-k}}={\displaystyle \frac{k^2-4ah}{4a^2+k^2}}\\ & \Rightarrow y=k+{\displaystyle \frac{-k^3+4ahk}{4a^2+k^2}}\end{array}$$
By doing the LCM and adding the terms we get
$$\begin{array}{rl}& \Rightarrow y={\displaystyle \frac{4a^{2}k+k^3-k^3+4ahk}{4a^2+k^2}}\\ & \Rightarrow y={\displaystyle \frac{4ak(h+a)}{4a^2+k^2}}\end{array}$$
Therefore, the co – ordinates of point $${}^{\prime }{T}^{\prime }$$ are
$$\Rightarrow T=({\displaystyle \frac{h{k}^2+2a{k}^2-4a^{2}h}{4a^2+k^2}},{\displaystyle \frac{4ak(h+a)}{4a^2+k^2}})$$
Let us find the length of ‘ST’ using the distance formula
We know that the formula of distance between $$(x_1,y_1),(x_2,y_2)$$ is given as
$$ST=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$$
By using the above formula we get
$$\begin{array}{rl}& \Rightarrow ST=\sqrt{{({\displaystyle \frac{h{k}^2+2a{k}^2-4a^{2}h}{4a^2+k^2}}-a)}^2+{\left({\displaystyle \frac{4ak(h+a)}{4a^2+k^2}}\right)}^2}\\ & \Rightarrow ST={\displaystyle \frac{1}{4a^2+k^2}}\sqrt{{(h{k}^2+2a{k}^2-4a^{2}h-4a^3-a{k}^2)}^2+16a^2{k}^2{(h+a)}^2}\end{array}$$
By taking the common terms from the first term inside the square root we get
$$\begin{array}{rl}& \Rightarrow ST={\displaystyle \frac{1}{4a^2+k^2}}\sqrt{{(k-4a^2)}^2{(h+a)}^2+16a^2{k}^2{(h+a)}^2}\\ & \Rightarrow ST={\displaystyle \frac{(h+a)}{4a^2+k^2}}\sqrt{{(k-4a^2)}^2+16a^2{k}^2}\\ & \Rightarrow ST={\displaystyle \frac{(h+a)}{4a^2+k^2}}(4a^2+k^2)\\ & \Rightarrow ST=h+a\end{array}$$
Now let us find the distance of pint P to the directirx
We know that the equation of directrix is given as
$$\Rightarrow x+a=0$$
We know that formula of the distance of $$P(h,k)$$ to line $$ax+by+c=0$$ is given as
$$D={\displaystyle \frac{|ah+bk+c|}{\sqrt{a^2+b^2}}}$$
By using the above formula the distance of point $$P(h,k)$$ to $$x+a=0$$ is given as
$$\begin{array}{rl}& \Rightarrow D={\displaystyle \frac{|h+0+a|}{\sqrt{1^2+0^2}}}\\ & \Rightarrow D=h+a\end{array}$$
Here we can see that
$$\Rightarrow ST=D$$
Therefore we can say that the foot of this perpendicular from the focus is equal to distance of point P from its directrix.
Hence the required result has been proved.

Note: There is a shortcut for this solution.
The foot of perpendicular of point P on its polar with respect to curve $$S\equiv y^2=4ax$$ will be same as P. that means in the above solution the points $${}^{\prime }{P}^{\prime }$$ and $${}^{\prime }{T}^{\prime }$$ will coincide.
Therefore, if we can prove that $$D=SP$$ then we can easily prove the required result.
By using the distance formula the length of SP is
$$\Rightarrow SP=\sqrt{{(h-a)}^2+k^2}$$
Since point $$P(h,k)$$ lies on parabola we get
$$\Rightarrow k^2=4ah$$
By substituting the this formula in above equation we get
$$\begin{array}{rl}& \Rightarrow SP=\sqrt{{(h-a)}^2+4ah}\\ & \Rightarrow SP=h+a=D\end{array}$$
Hence the required result has been proved.