Standard form of Parabola x$^{2}$= -4ay

We will discuss about the standard form of parabola x$^{2}$ = -4ay

Equation y$^{2}$ = -4ax (a > 0) represents the equation of a parabola whose co-ordinate of the vertex is at (0, 0), the co-ordinates of the focus are (0, -a), the equation of directrix is y = a or y - a = 0, the equation of the axis is x = 0, the axis is along negative y-axis, the length of its latus rectum = 4a and the distance between its vertex and focus is a.

$Standard form of Parabola x^2= -4ay$ Standard form of Parabola x^2= -4ay

Solved examples based on the standard form of parabola x$^{2}$ = -4ay:

1. Find the axis, co-ordinates of vertex and focus, length of latus rectum and the equation of directrix of the parabola x$^{2}$ = -16y

Solution:

The given parabola x$^{2}$ = -16y

⇒ x$^{2}$ = -4 ∙ 4 y

Compare the above equation with standard form of parabola x$^{2}$ = -4ay, we get, a = 4.

Therefore, the axis of the given parabola is along negative y-axis and its equation is x = 0

The co-ordinates of its vertex are (0, 0) and the co-ordinates of its focus are (0, -4); the length of its latus rectum = 4a = 4 ∙ 4 = 16 units and the equation of its directrix is y = a i.e., y = 4 i.e., y - 4 = 0.

2. Find the axis, co-ordinates of vertex and focus, length of latus rectum and the equation of directrix of the parabola 3x$^{2}$ = -8y

Solution:

The given parabola 3x$^{2}$ = -8y

⇒ x$^{2}$ = -$\frac{8}{3}$y

⇒ x$^{2}$ = -4 ∙ $\frac{2}{3}$ y

Compare the above equation with standard form of parabola x$^{2}$ = -4ay, we get, a = $\frac{2}{3}$.

Therefore, the axis of the given parabola is along negative y-axis and its equation is x = 0

The co-ordinates of its vertex are (0, 0) and the co-ordinates of its focus are (0, -$\frac{2}{3}$); the length of its latus rectum = 4a = 4 ∙ $\frac{2}{3}$ = $\frac{8}{3}$ units and the equation of its directrix is y = $\frac{2}{3}$ i.e., 3y = 2 i.e., 3y - 2 = 0.

● The Parabola