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Step-by-Step Solution
Step 1: Write the given parabola and identify its focus
The equation of the given parabola is
$y^2 = 4ax.$
For a parabola of this form, its focus is at
$S(a, 0).$
Step 2: Choose a general point on the parabola
A standard parametric point on the parabola
$y^2 = 4ax$
can be taken as
$P(at^2, 2at).$
Step 3: Find the midpoint of the segment joining S and P
Let $M(h, k)$ be the midpoint of $SP.$ Then
$h = \frac{a t^2 + a}{2}, \quad k = \frac{2 a t + 0}{2}.$
Simplifying gives:
$h = \frac{a(t^2 + 1)}{2}, \quad k = a t.$
Step 4: Express t in terms of h and k
From $k = a t,$ we have
$t = \frac{k}{a}.$
Also, from
$h = \frac{a(t^2 + 1)}{2},$
we get
$2h = a(t^2 + 1).$
Hence,
$t^2 = \frac{2h}{a} - 1.$
Substituting $t^2 = \left(\frac{k}{a}\right)^2 = \frac{k^2}{a^2}$, we obtain:
$\frac{k^2}{a^2} = \frac{2h}{a} - 1.$
Step 5: Derive the locus in terms of x and y
Multiply throughout by $a^2$:
$k^2 = a^2 \left( \frac{2h}{a} - 1 \right) = a(2h) - a^2.$
Since $h$ and $k$ represent $x$ and $y$ respectively, rename $(h, k)$ as $(x, y)$:
$y^2 = a(2x) - a^2.$
Therefore, the locus is:
$y^2 = 2a \left(x - \frac{a}{2}\right).$
Step 6: Identify the directrix of the new parabola
The standard form
$y^2 = 4A (x - x_0)$
has a directrix given by
$x - x_0 = -A.$
In our equation
$y^2 = 2a \left(x - \frac{a}{2}\right),$
we can see $A = \frac{2a}{4} = \frac{a}{2}$ and $x_0 = \frac{a}{2}.$ Thus the directrix is:
$x - \frac{a}{2} = -\frac{a}{2}.$
This simplifies to
$x = 0.$
Final Answer
The directrix of the new parabola is
$x = 0.$
