Answer to Question #224755 in Analytic Geometry for anuj

Question

The normal to the parabola y²= 4ax at the point P(at² , 2at) meets the x-axis at A. Find the equation of the locus of the midpoint of AP as t varies.

Answer 1

To find the normal;

At A(y,x) but y=0

Since;

y²=4ax

$2y \frac{dy}{dx} =4a$

$\frac{dy}{dx} = \frac{2a}{y}$

Slope of the tangent at P(at² , 2at); $\frac{dy}{dx} =\frac{2a}{2at} =\frac{a}{t}$

Hence the slope at normal P is given as $\frac{-1}{\frac{1}{t}} =-t$

Therefore equation of the normal at P is given by $tx+y=2at+at^3$

Hence;

$y-2at=-t(x-at^2)$

But y=0

$−2at=−tx+at^3 \\ Rewrite \,as \\ tx=at^3+2at\\ Divide \,by \,t,we \,get\\ x=at^2+2ax=at^2 +2a$

Take the midpoint of AP to be B(h,k)

Here

$h=\frac{at^2 +2a +at^2}{2} =at^2 +a$

$k= \frac{0+2at}{2}= at$

since; y $^2 =4ax$

We have the equation of the locus as t varies given by

$(2at)^2 =4a(at^2)$

Using h and k ,we can make the following substitutions

$(2 x k)^2 =4a(h-a )$

$4k^2 =4ah-4a^2$

Rewrite as

$k^2 =ah -a^2$

$k^2 - ah +a^2 =0$

Write in terms of x and y as

$y^2-ax -a^2=0$