Find the vertex, axis, focus and directrix of the parabola $\;$ $\left(y + 3\right)^2 = 4x + 4$.
Equation of given parabola: $\;$ $\left(y + 3\right)^2 = 4x + 4$
i.e. $\;$ $\left(y + 3\right)^2 = 4 \left(x + 1\right)$ $\;\;\; \cdots \; (1)$
To shift the origin to the point $\left(-1, -3\right)$,
let $\;$ $x + 1 = X$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y + 3 = Y$ $\;\;\; \cdots \; (2b)$
In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes
$Y^2 = 4 X$ $\;\;\; \cdots \; (3)$
- Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$
i.e. $\;$ $X = 0, \;\;\; Y = 0$
When $\;$ $X = 0$
$\implies$ $x + 1 = 0$ $\;$ i.e. $\;$ $x = -1$ $\;\;\;$ [by equation $(2a)$]
When $\;$ $Y = 0$
$\implies$ $y + 3 = 0$ $\;$ i.e. $\;$ $y = -3$ $\;\;\;$ [by equation $(2b)$]
$\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(-1, -3\right)$
- Axis of equation $(3)$ is $\;\;$ $Y = 0$
$\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $y + 3 = 0$ $\;\;\;$ [by equation $(2b)$]
i.e. $\;$ $y = -3$
- Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $Y^2 = 4a X$ $\;\;$ gives
$a = 1$
$\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(a, 0\right) = \left(1, 0\right)$
$\implies$ $X = 1, \;\;\; Y = 0$
When $\;$ $X = 1$
$\implies$ $x + 1 = 1$ $\;$ i.e. $\;$ $x = 0$ $\;\;\;$ [by equation $(2a)$]
When $\;$ $Y = 0$
$\implies$ $y + 3 = 0$ $\;$ i.e. $\;$ $y = -3$ $\;\;\;$ [by equation $(2b)$]
$\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(0, -3\right)$
- Directrix of equation $(3)$ is $\;\;\;$ $X = -a$
i.e. $\;$ $X = -1$
i.e. $x + 1 = -1$ $\;\;\;$ [by equation $(2a)$]
$\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $x + 2 = 0$