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

Solution:

To find the normal;

At A(y,x) but y=0

Since;

$y^2=4ax$

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

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

Slope of the tangent at

$P(at^2,2at)$ ;$m_1=\frac{dy}{dx}=\frac{2a}{2at}$ =$\frac1t$

Hence slope of the normal at P is given as;

$m_2=\dfrac{-1}{\dfrac1t}$ =$-t$

Therefore equation of the normal is given by;

$y=m_2x+c \\\Rightarrow y=-tx+c$

Put $x=at^2,y=2at$

So, $c=2at+at^3$

Thus, $tx+y=2at+at^3$

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;

$x=at^2+2a$

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 at P;

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

Using h and k ,we can make the following substitutions;

$(2×k)^2=4a(h-a)$

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

Rewrite as ;

$k^2=ah-a^2$

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

Writing it in terms of x and y as;

$y^2-ax+a^2=0$ which is the standard locus at midpoint AP.