Find the equation of the parabola whose vertex is at $\left(3, -2\right)$ and the focus at $\left(6, 2\right)$.
Given: $\;$ Vertex $= V = \left(3, -2\right)$, $\;\;\;$ Focus $= F = \left(6, 2\right)$
The axis is the line joining the vertex and the focus.
$\therefore \;$ Slope of axis $= \dfrac{2 + 2}{6 - 3} = \dfrac{4}{3}$
The directrix is the line perpendicular to the axis and passing through the vertex.
$\therefore \;$ Slope of directrix $= \dfrac{-3}{4}$
Let $\;$ $Z \left(x_1, y_1\right)$ $\;$ be the point of intersection of the axis and the directrix.
Then, vertex $\;$ $V \left(3, -2\right)$ $\;$ is the midpoint of the segment $\;$ $ZF$ $\;$ where $\;$ $Z = \left(x_1, y_1\right)$ $\;$ and $\;$ $F = \left(6, 2\right)$
$\therefore \;$ By midpoint formula,
$3 = \dfrac{x_1 + 6}{2}$ $\implies$ $x_1 = 0$
and $\;$ $-2 = \dfrac{y_1 + 2}{2}$ $\implies$ $y_1 = -6$
$\therefore \;$ $Z \left(x_1, y_1\right) = \left(0, -6\right)$
$\therefore \;$ Equation of directrix $\;$ (with slope $\dfrac{-3}{4}$ and passing through $Z$) $\;$ is
$y + 6 = \dfrac{-3}{4} \left(x - 0\right)$
i.e. $\;$ $3x + 4y + 24 = 0$
Let $\;$ $P \left(x, y\right)$ $\;$ be any point on the required parabola.
Then, by definition,
distance of $P$ from focus $=$ distance of $P$ from the directrix
i.e. $\;$ $\sqrt{\left(x - 6\right)^2 + \left(y - 2\right)^2} = \left|\dfrac{3x + 4y + 24}{\sqrt{3^2 + 4^2}}\right|$
i.e. $\;$ $x^2 + y^2 - 12x - 4y + 36 + 4 = \dfrac{9x^2 + 16y^2 + 576 + 24xy + 144x + 192y}{5}$
i.e. $\;$ $5x^2 + 5y^2 - 60x - 20y + 200 = 9x^2 + 16y^2 + 24xy + 144x + 192y + 576$
i.e. $\;$ $4x^2 + 11y^2 - 24xy + 204x + 212y + 376 = 0$ $\;\;\; \cdots \; (1)$
Equation $(1)$ is the equation of the required parabola.