Before we begin, note that in the given equation the variable that is raised to the

second

power

is

x .

x^{2} = 3 y

Let's recall the general form of the equation for this type of parabola.

y = \frac{1}{4 p} (x - h)^{2} + k

Let's rearrange the given equation.

x^{2} = 3 y

$LHS \cdot \frac{1}{3} = RHS \cdot \frac{1}{3}$

\frac{1}{3} x^{2} = y

Rearrange equation

y = \frac{1}{3} x^{2}

We need to identify the values of

k,

h,

and

p .

Let's start with

p .

To do so, we will solve the equation

\frac{1}{4 p} = \frac{1}{3} .

We set it equal to

\frac{1}{3}

because it is the coefficient of the

x^{2}

term.

\frac{1}{4 p} = \frac{1}{3}

▼

Solve for

p

$LHS \cdot 3 = RHS \cdot 3$

\frac{1}{4 p} (3) = 1

$\frac{1}{b} \cdot a = \frac{a}{b}$

\frac{3}{4 p} = 1

$LHS \cdot p = RHS \cdot p$

\frac{3}{4} = p

Rearrange equation

p = \frac{3}{4}

Knowing that

p = \frac{3}{4},

we can rewrite the equation.

y = \frac{1}{3} x^{2} ⇕ y = \frac{1}{4 ( \frac{3}{4} )} (x - 0)^{2} + 0

Now we have that

k = 0,

h = 0,

and

p = \frac{3}{4} .

By recalling the corresponding formulas, we can find the focus, directrix, and axis of symmetry of the parabola.

	Focus	Directrix	Axis of Symmetry
Formula	$(h, k + p)$	$y = k - p$	$x = h$
Value	$(0, 0 + \frac{3}{4})$ $⇓$ $(0, \frac{3}{4})$	$y = 0 - \frac{3}{4}$ $⇓$ $y = - \frac{3}{4}$	$x = 0$

Now, let's draw the parabola using the obtained information.

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