Find the equation of the parabola whose focus is at $\left(-2, -1\right)$ and the latus rectum joins the points $\left(-2, 2\right)$ and $\left(-2, -4\right)$.
Given: $\;$ Latus rectum joins the points $\;$ $\left(-2, 2\right)$ $\;$ and $\;$ $\left(-2, -4\right)$
Length of latus rectus $= \sqrt{\left(-2 + 2\right)^2 + \left(2 + 4\right)^2} = 6$
But length of latus rectum $= 4a$
i.e. $\;$ $4a = 6$ $\implies$ $a = \dfrac{6}{4} = \dfrac{3}{2}$
Given: $\;$ Focus $= \left(-2, -1\right)$
Let the vertex of the required parabola be $= \left(\alpha, -1\right)$
Then,
Distance between focus and vertex $= a$
i.e. $\;$ $\sqrt{\left(\alpha + 2\right)^2 + \left(-1 + 1\right)^2} = \dfrac{3}{2}$
i.e. $\;$ $\alpha + 2 = \dfrac{3}{2}$ $\implies$ $\alpha = \dfrac{-1}{2}$
$\therefore \;$ The vertex of the required parabola is $= \left(h, k\right) = \left(\dfrac{-1}{2}, -1\right)$
Let the equation of the required parabola be
$\left(y - k\right)^2 = \pm 4 a \left(x - h\right)$
i.e. $\;$ $\left(y + 1\right)^2 = \pm 6 \left(x + \dfrac{1}{2}\right)$
i.e. $\;$ $y^2 + 2y + 1 = 6x + 3$ $\;\;$ or $\;\;$ $y^2 + 2y + 1 = -6x - 3$
i.e. $\;$ $y^2 + 2y - 6x -2 = 0$ $\;\;\; \cdots \; (1a)$
or $\;$ $y^2 + 2y + 6x + 4 = 0$ $\;\;\; \cdots \; (1b)$
Equations $(1a)$ and $(1b)$ are the required equations of the parabola.