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

second

power

is

y .

y^{2} = \frac{2}{5} x

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

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

Let's rearrange the given equation.

y^{2} = \frac{2}{5} x

$LHS \cdot \frac{5}{2} = RHS \cdot \frac{5}{2}$

\frac{5}{2} y^{2} = x

Rearrange equation

x = \frac{5}{2} y^{2}

We need to identify the values of

h,

k,

and

p .

Let's start with

p .

To do so, we will solve the equation

\frac{1}{4 p} = \frac{5}{2} .

We set it equal to

\frac{5}{2}

because it is the coefficient of the

y^{2}

term.

\frac{1}{4 p} = \frac{5}{2}

▼

Solve for

p

$LHS \cdot \frac{2}{5} = RHS \cdot \frac{2}{5}$

\frac{1}{4 p} (\frac{2}{5}) = 1

Multiply fractions

\frac{2}{2 0 p} = 1

$LHS \cdot p = RHS \cdot p$

\frac{2}{2 0} = p

$\frac{a}{b} = \frac{a / 2}{b / 2}$

\frac{1}{1 0} = p

Rearrange equation

p = \frac{1}{1 0}

Knowing that

p = \frac{1}{1 0},

we can rewrite the equation.

x = \frac{5}{2} y^{2} ⇕ x = \frac{1}{4 ( \frac{1}{1 0} )} (y - 0)^{2} + 0

Now we have that

h = 0,

k = 0,

and

p = \frac{1}{1 0} .

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

	Focus	Directrix	Axis of Symmetry
Formula	$(h + p, k)$	$x = h - p$	$y = k$
Value	$(0 + \frac{1}{1 0}, 0)$ $⇓$ $(\frac{1}{1 0}, 0)$	$x = 0 - \frac{1}{1 0}$ $⇓$ $x = - \frac{1}{1 0}$	$y = 0$

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

{{ 'ml-heading-exercise' | message }} {{ exercise.name }} {{ 'ml-label-page' | message }} {{ exercise.page }}