show that the locus of the middle points of chords of parabola $y^2 = 4ax$ passing throughthe vertex is a parabola?

Question

show that the locus of the middle points of chords of parabola $$y^2 = 4ax$$ passing through the vertex is a parabola?

Solution

the Vertex is $O(0.0)$, which is one end of the chord. Let the other end be a varaible point $P$ given by $(at^2,2at)$.

Let $M(p,q)$ be the midpoint of the chord $OP$. Midpoint of $OP$ is $(at^2/2,at)$.

So,

$$p = \frac{at^2}{2}$$ and $$q = at$$

Now we have to eliminate $t$ and get the relation between $p$ and $q$ to get the locus.

So,

$$t = \frac{q}{a}$$

Substitute this in the equation of $p$, and we will get:

$$p = \frac{a}{2}\left(\frac{q}{a}\right)^2$$

So we have, $$q^2 = 2ap$$

Which is a parabola of the form $y^2 = 2ax$

My question is : what is the variable $t$ , why it has been introduced in the solution, and how the coordinate of point $P(at^2,2at)$ have been obtained?

can this problem be solved without using parametric equations of parabola since it is not included in the curriculum?

Answer 1

From the equation $y^2=4ax$ we solve for $x$ and we get $x=\dfrac{y^2}{4a}$ means that the points of the parabola are all in the form $P\left(\dfrac{y^2}{4a},\;y\right)$.

A chord $OP$ has midpoint $M\left(\dfrac{y^2}{8a},\;\dfrac{y}{2}\right)$ which means that $y_M^2=\dfrac{y^2}{4}$ that is $y^2=4y_M^2$

In a similar way we show that $x_M=\dfrac{y^2}{8a}$ then $y^2=8ax_M$

So we can conclude that $4y_M^2=8ax_M$ and finally $y_M^2=2ax_M$

as $P$ runs on to the parabola, $M$ runs on the curve $y^2=2ax$

I hope it is clear. (Actually I have used hidden parametrization)

Answer 2

Let $AB$ be our chord and $AB\perp$ $x$-axes.

Thus, a midpoint of $AB$ placed on the x-axes and since the vertex placed on $x$-axes, we are done!

If our chords are parallel then it's another problem!