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.

$x=2y^2$

Short answer

a. The parabola $x=2y^2$ has its focus at $\left(\frac{1}{8},0\right)$ . Its focal diameter is $\frac{1}{2}$ and directrix is $x=-\frac{1}{8}$.

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

Step-by-step solution

a.Step 1. Given information.

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

Step 2. Write the concept.

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

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

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

$\frac{1}{2}x=\frac{2}{2}{y}^2$

From the above equation $4a=\frac{1}{2}$, so the focal diameter is $\frac{1}{2}$

By solving $a=\frac{1}{8}$ for a,

We get $a=\frac{1}{8}$

$\therefore$ the focus is $\left(\frac{1}{8},0\right)$ and the directrix is $x=-\frac{1}{8}$

b.Step 1. Given information.

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

Step 2. Write the concept.

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

Converting the equation in standard form

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

$x=2y^2$

Hence the coordinates of the focus will be $\left(\frac{1}{8},0\right)$

And the equation of the directrix will be $x=-\frac{1}{8}$