What is an equation of the parabola with vertex at the origin and focus (-4, 0)?

Answer 1

Since the vertex is at the origin, the equation of the parabola is of the form $y=ax^2$. The focus is $4$ units to the left of the vertex, so the directrix is the vertical line $x=4$. This means that the distance from any point on the parabola to $(4,0)$ is equal to its distance to the vertex $(0,0)$.

Using the distance formula, we can write an equation for this condition:

$$\sqrt{(x-4)^2+y^2}= \sqrt{x^2+y^2}$$

Squaring both sides:

$$(x-4)^2+y^2=x^2+y^2$$

Expanding the left side:

$$x^2-8x+16+y^2=x^2+y^2$$

Simplifying:

$$-8x+16=0$$

Therefore, $x=2$. Substituting into the equation, we find:

$$y=a(2)^2=4a$$

Since the focus is on the negative x-axis, this means the parabola opens to the left, so $a<0$.

Thus, the equation of the parabola is $\boxed{y=-\frac{1}{4}x^2}$.