Browse Questions

# Find the co-ordinates of the focus, axis of the parabola, equations of the directrix and length of latus rectum of the parabola $x^2=-16y$

$\begin{array}{1 1} (0,-4) \quad \text{y-axis} \quad y=4 \quad 16 \\ (0,4) \quad \text{y-axis} \quad y=4 \quad 8 \\ (0,-4) \quad \text{y-axis} \quad y=2\quad 8 \\ (0,4) \quad \text{x-axis} \quad y=4 \quad 16\end{array}$

Can you answer this question?

Toolbox:
Given $x^2=-16y$
The co-efficient of $y$ is negative, The parabola opens downward. $x^2=16y \rightarrow -4a = -16 \rightarrow a = 4$
1) Therefore co-ordinates of the focus = $(0,-a) = (0,-4)$
2) Since $x^2=-16y$, the axis of the parabola is the y-axis
3) The Equation of the directrix $y = a \rightarrow y = 4$
4) The Length of the Latus Rectum $= 4a =4 \times 4 = 16$
answered Apr 1, 2014
edited Apr 1, 2014