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.

Short answer

a. The parabola $x^2+12y=0$has its focus at $\left(0,-3\right)$. Its focal diameter is 12 and directrix is $y=3$.

b. The graph of parabola $x+\frac{1}{5}{y}^2=0$is

Step-by-step solution

a.Step 1. Given information.

The given equation of a parabola $x^2+12y=0$.

Step 2. Write the concept.

The given equation of a parabola $x^2+12y=0$

Putting the equation in standard form $x^2=4ay$

Thus we get, $x^2+12y=0$

$x^2=-4\left(3\right)y$

From the above equation $4a=-12$, so the focal diameter is 12

By solving $a=-3$ for a,

We get $a=-3$

$\therefore$ the focus is $\left(0,-3\right)$ and the directrix is$y=3$

b.Step 1. Given information.

The given equation of a parabola $x^2+12y=0$.

Step 2. Write the concept.

The given equation of a parabola $x^2+12y=0$.

Converting the equation in standard form

Thus, $x^2+12y=0$ which represents a downward parabola with vertex at (0,0)

$x^2+12y=0$

Hence the coordinates of the focus will be $\left(0,-3\right)$

And the equation of the directrix will be $y=3$