Coordinate Geometry - Parabola

Find the vertex, axis, focus and directrix of the parabola $\;$ $x^2 + 2gx + 2fy + c = 0$.


Equation of given parabola: $\;$ $x^2 + 2gx + 2fy +c = 0$

i.e. $\;$ $\left(x^2 + 2gx+ g^2\right) + 2fy + c - g^2 = 0$

i.e. $\;$ $\left(x + g\right)^2 = -2fy + g^2 - c$

i.e. $\;$ $\left(x + g\right)^2 = -2f \left(y - \dfrac{g^2 - c}{2f}\right)$ $\;\;\; \cdots \; (1)$

To shift the origin to the point $\left(-g, \dfrac{g^2 - c}{2f}\right)$,

let $\;$ $x + g = X$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $y - \dfrac{g^2 - c}{2f} = Y$ $\;\;\; \cdots \; (2b)$

In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$X^2 = -2fY$ $\;\;\; \cdots \; (3)$

  1. Vertex of equation $(3)$ is $\;$ $\left(0, 0\right)$

    i.e. $\;$ $X = 0, \;\;\; Y = 0$

    When $\;$ $X = 0$

    $\implies$ $x + g = 0$ $\;$ i.e. $\;$ $x = -g$ $\;\;\;$ [by equation $(2a)$]

    When $\;$ $Y = 0$

    $\implies$ $y - \dfrac{g^2 - c}{2f} = 0$ $\;$ i.e. $\;$ $y = \dfrac{g^2 - c}{2f}$ $\;\;\;$ [by equation $(2b)$]

    $\therefore \;$ Vertex of equation $(1)$ is $\;$ $\left(-g, \dfrac{g^2 - c}{2f}\right)$

  2. Axis of equation $(3)$ is $\;\;$ $X = 0$

    $\therefore \;$ Axis of equation $(1)$ is $\;\;\;$ $x + g = 0$ $\;\;\;$ [by equation $(2a)$]

    i.e. $\;$ $x = -g$

  3. Comparing equation $(3)$ with the standard equation of parabola $\;\;$ $X^2 = -4a Y$ $\;\;$ gives

    $4a = 2f$ $\implies$ $a = \dfrac{f}{2}$

    $\therefore \;$ Focus of equation $(3)$ is $\;\;\;$ $\left(0, -a\right) = \left(0, \dfrac{-f}{2}\right)$

    $\implies$ $X = 0, \;\;\; Y = \dfrac{-f}{2}$

    When $\;$ $X = 0$

    $\implies$ $x + g = 0$ $\;$ i.e. $\;$ $x = -g$ $\;\;\;$ [by equation $(2a)$]

    When $\;$ $Y = \dfrac{-f}{2}$

    $\implies$ $y - \dfrac{g^2 - c}{2f} = \dfrac{-f}{2}$ $\;$ i.e. $\;$ $y = \dfrac{g^2 - f^2 - c}{2f}$ $\;\;\;$ [by equation $(2b)$]

    $\therefore \;$ Focus of equation $(1)$ is $\;\;\;$ $\left(-g, \dfrac{g^2 - f^2 - c}{2f}\right)$

  4. Directrix of equation $(3)$ is $\;\;\;$ $Y = a$

    i.e. $\;$ $Y = \dfrac{f}{2}$

    i.e. $y - \dfrac{g^2 - c}{2f} = \dfrac{f}{2}$ $\;\;\;$ [by equation $(2b)$]

    i.e. $\;$ $y= \dfrac{g^2 - f^2 - c}{2f}$

    $\therefore \;$ Directrix of equation $(1)$ is $\;\;\;$ $y = \dfrac{g^2 - f^2 - c}{2f}$