Find the vertex, axis, focus and directrix of the parabola $\;$ $\left(x - 4\right)^2 = -8y + 5$.
Equation of given parabola: $\;$ $\left(x - 4\right)^2 = -8y + 5$
i.e. $\;$ $\left(x - 4\right)^2 = -8 \left(y - \dfrac{5}{8}\right)$ $\;\;\; \cdots \; (1)$
To shift the origin to the point $\left(4, \dfrac{5}{8}\right)$,
let $\;$ $x - 4 = X$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - \dfrac{5}{8} = Y$ $\;\;\; \cdots \; (2b)$
In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes
$X^2 = -8Y$ $\;\;\; \cdots \; (3)$
- Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$
i.e. $\;$ $X = 0, \;\;\; Y = 0$
When $\;$ $X = 0$
$\implies$ $x - 4 = 0$ $\;$ i.e. $\;$ $x = 4$ $\;\;\;$ [by equation $(2a)$]
When $\;$ $Y = 0$
$\implies$ $y - \dfrac{5}{8} = 0$ $\;$ i.e. $\;$ $y = \dfrac{5}{8}$ $\;\;\;$ [by equation $(2b)$]
$\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(4, \dfrac{5}{8}\right)$
- Axis of equation $(3)$ is $\;\;$ $X = 0$
$\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $x - 4 = 0$ $\;\;\;$ [by equation $(2a)$]
i.e. $\;$ $x = 4$
- Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $X^2 = -4a Y$ $\;\;$ gives
$4a = 8$ $\implies$ $a = 2$
$\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(0, -a\right) = \left(0, -2\right)$
$\implies$ $X = 0, \;\;\; Y = -2$
When $\;$ $X = 0$
$\implies$ $x - 4 = 0$ $\;$ i.e. $\;$ $x = 4$ $\;\;\;$ [by equation $(2a)$]
When $\;$ $Y = -2$
$\implies$ $y - \dfrac{5}{8} = -2$ $\;$ i.e. $\;$ $y = \dfrac{-11}{8}$ $\;\;\;$ [by equation $(2b)$]
$\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(4, \dfrac{-11}{8}\right)$
- Directrix of equation $(3)$ is $\;\;\;$ $Y = a$
i.e. $\;$ $Y = 2$
i.e. $y - \dfrac{5}{8} = 2$ $\;\;\;$ [by equation $(2b)$]
i.e. $\;$ $y= \dfrac{21}{8}$
$\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $8y = 21$