(a)
We have:
\displaystyle 2y \frac {\mathrm dy} {\mathrm dx} = 4a
So:
\displaystyle \frac {\mathrm dy} {\mathrm dx} = \frac {2a} y
At y = 2aq we have:
\displaystyle \frac {\mathrm dy} {\mathrm dx} = \frac {2a} {2aq} = \frac 1 q
So the tangent to C at Q has equation:
\displaystyle y - 2aq = \frac 1 q (x - aq^2)
So:
qy - 2aq^2 = x - aq^2
so:
qy = x + aq^2
(b)
If this tangent passes through X then:
\displaystyle q \times 0 = -\frac 1 4 a + aq^2
So, since a > 0:
\displaystyle q^2 = \frac 1 4
since q > 0, we have:
\displaystyle q = \frac 1 2
The directrix of C is the line x = -a.
We therefore have, at this point:
\displaystyle \frac 1 2 y = -a + a\left(\frac 1 2\right)^2
So:
\displaystyle y = -2 + \frac 2 4 a = -\frac 3 2 a
So \displaystyle D = \left(-a, -\frac 3 2 a\right)
©
We want to find the area of the triangle with vertices F = (a, 0), \displaystyle X = \left(-\frac 1 4 a, 0\right), \displaystyle D = \left(-a, -\frac 3 2 a\right). Note that this has base length \displaystyle a - \left(-\frac 1 4 a\right) = \frac 5 4 a, and perpendicular height \dfrac 3 2 a, so the area of FXD is:
\displaystyle \frac 1 2 \times \frac 5 4 a \times \frac 3 2 a = \frac {15} {16}a^2
1 Like