Question

The equation of the parabola whose vertex is (a, 0) and the directrix has the equation x + y = 3a, is
(a) x² + y² + 2xy + 6ax + 10ay + 7a² = 0
(b) x² − 2xy + y² + 6ax + 10ay − 7a² = 0
(c) x² − 2xy + y² − 6ax + 10ay − 7a² = 0
(d) none of these

Answer 1

(b) x² − 2xy + y² + 6ax + 10ay − 7a² = 0

Given:
The vertex is at (a, 0) and the directrix is the line x + y = 3a.

The slope of the line perpendicular to x + y = 3a is 1.

The axis of the parabola is perpendicular to the directrix and passes through the vertex.

∴ Equation of the axis of the parabola = $y-0=1\left(x-a\right)$ (1)

Intersection point of the directrix and the axis is the intersection point of (1) and x + y = 3a.

Let the intersection point be K.

Therefore, the coordinates of K are $\left(2a,a\right)$.

The vertex is the mid-point of the segment joining K and the focus (h, k).
∴ $$\begin{gathered} a=\frac{2a+h}{2},0=\frac{a+k}{2} \\ h=0,k=-a \end{gathered}$$

Let P (x, y) be any point on the parabola whose focus is S (h, k) and the directrix is x + y = 3a.

Draw PM perpendicular to x + y = 3a.
Then, we have:
$$\begin{gathered} SP=PM \\ \Rightarrow S{P}^2=P{M}^2 \\ \Rightarrow {\left(x-0\right)}^2+{\left(y+a\right)}^2={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^2 \\ \Rightarrow x^2+{\left(y+a\right)}^2={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^2 \\ \Rightarrow 2x^2+2y^2+2a^2+4ay=x^2+y^2+9a^2+2xy-6ax-6ay \\ \Rightarrow x^2+y^2-7a^2+10ay+6ax-2xy=0 \end{gathered}$$