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 $9x=y^2$has its focus at $9x=y^2$. Its focal diameter is 9 and directrix is $x=\frac{-9}{4}$.

b. The graph of parabola $9x=y^2$is

Step-by-step solution

a.Step 1. Given information.

The given equation of a parabola $9x=y^2$.

Step 2. Write the concept.

The given equation of a parabola $9x=y^2$

Putting the equation in standard form $y^2=4ax$

Thus we get, $9x=y^2$

$y^2=4\left(\frac{9}{4}\right)x$

From the above equation $4a=9$, so the focal diameter is 9

By solving $a=\frac{9}{4}$ for a,

We get $a=\frac{9}{4}$

$\therefore$the focus is $\left(\frac{9}{4},0\right)$and the directrix is $x=\frac{-9}{4}$

b.Step 1. Given information.

The given equation of a parabola $9x=y^2$.

Step 2. Write the concept.

The given equation of a parabola $9x=y^2$.

Converting the equation in standard form

Thus, $9x=y^2$ which represents a right facing parabola with vertex at (0,0)

$9x=y^2$

Hence the coordinates of the focus will be $\left(\frac{9}{4},0\right)$

And the equation of the directrix will be $x=\frac{-9}{4}$