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 $x^2=6y$

$\begin{array}{1 1}(0,\large\frac{3}{2}) \quad \text{y-axis} \quad y = \large\frac{-3}{2} \quad 6 \\ (0,\large\frac{3}{2}) \quad \text{x-axis} \quad y = \large\frac{-3}{2} \quad 6 \\ (0,\large\frac{-3}{2}) \quad \text{y-axis} \quad y = \large\frac{-1}{2} \quad 6 \\(0,\large\frac{-3}{2}) \quad \text{x-axis} \quad y = \large\frac{-3}{2} \quad 6 \end{array}$

Toolbox:
Given $x^2=6y$
The co-efficient of $y$ is positive, comparing with $x^2 = 4ay \rightarrow 4a = 6 \rightarrow a = \large\frac{3}{2}$
1) Therefore co-ordinates of the focus = $(0,a) = (0,\large\frac{3}{2})$
2) Since $x^2 = 6x$, the axis of the parabola is the y-axis
3) The Equation of the directrix $y = -a \rightarrow y = \large\frac{-3}{2}$
4) The Length of the Latus Rectum $= 4a = 4 \times \large\frac{3}{2}$ = 6