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 $5y=x^2$has its focus at $\left(0,\frac{5}{4}\right)$. Its focal diameter is 5 and directrix is $y=\frac{-5}{4}$.

b. The graph of parabola $5y=x^2$is

Step-by-step solution

a.Step 1. Given information.

The given equation of a parabola $5y=x^2$.

Step 2. Write the concept.

The given equation of a parabola $5y=x^2$

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

Thus we get, $5y=x^2$

$x^2=4\left(\frac{5}{4}\right)y$

From the above equation $4a=5$, so the focal diameter is $5$

By solving $a=\frac{5}{4}$ for a,

We get $a=\frac{5}{4}$

$\therefore$ the focus is $\left(0,\frac{5}{4}\right)$ and the directrix is $y=\frac{-5}{4}$

b.Step 1. Given information.

The given equation of a parabola $5y=x^2$.

Step 2. Write the concept.

The given equation of a parabola $5y=x^2$.

Converting the equation in standard form

Thus, $5y=x^2$ which represents a upward facing parabola with vertex at (0,0)

$5y=x^2$

Hence the coordinates of the focus will be $\left(0,\frac{5}{4}\right)$

And the equation of the directrix will be $y=\frac{-5}{4}$