Question

An equation of a parabola is

given.

(a) Find the focus, directrix, and focal diameter of the

parabola.

(b) Sketch a graph of the parabola and its directrix.

$x^2=8y$

Short answer

a. The parabola $x^2=-4y$has its focus at $\left(0,2\right)$. Its focal diameter is 8 and directrix is $y=-2$.

b. The graph of parabola $x^2=8y$is

Step-by-step solution

a.Step 1. Given information.

The given equation of a parabola $x^2=8y$.

Step 2. Write the concept.

The given equation of a parabola $x^2=8y$

Putting the equation in standard form $x^2=8y$

Thus we get, $x^2=8y$

$x^2=4\left(2\right)y$

From the above equation $4a=8$, so the focal diameter is 8 By solving $a=2$ for a,

We get $a=2$

$\therefore$ the focus is $\left(0,2\right)$ and the directrix is $y=-2$

b.Step 1. Given information.

The given equation of a parabola $x^2=8y$.

Step 2. Write the concept.

The given equation of a parabola $x^2=8y$.

Converting the equation in standard form

Thus, $x^2=8y$ which represents a upward facing parabola with vertex at (0,0)

$x^2=8y$

Hence the coordinates of the focus will be $\left(0,2\right)$

And the equation of the directrix will be $y=-2$