Back to Exercise: Derive parabola equation

Exercises: Derive the Equation of a Parabola Given Focus and Directrix

Show your work for each problem. Unless a problem says otherwise, assume the axis of symmetry is horizontal or vertical, and give equations in standard form.

Grade 10·23 problems·~35 min·Common Core Math - HS Geometry·standard·hsg-gpe-a-2

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Warm-Up: Review What You Know

These problems review skills you have already learned.

Use the distance formula to find the distance between $(3, 7)$ and $(-1, 4)$ .

How far is the point $(6, 2)$ from the horizontal line $y = -3$ ?

$\sqrt{45}$ , found by applying the distance formula from $(6, 2)$ to the point $(0, -3)$ where the line crosses the $y$ -axis.

$5$ , because the shortest path from a point to a horizontal line is straight down, a change of $2 - (-3)$ units in $y$ only.

$3$ , because the line is $y = -3$ and the distance to a line named by a number is that number.

$\sqrt{85}$ , found by applying the distance formula from $(6, 2)$ to the origin-side point $(-3, 0)$ on the line.

Expand $(y + 4)^2$ .

$y^2 + 16$ , since squaring a sum squares each term.

$y^2 + 4y + 16$ , using the first term, the coefficient $4$ , and the square of $4$ .

$y^2 + 8y + 16$ , using $(a + b)^2 = a^2 + 2ab + b^2$ with $a = y$ and $b = 4$ .

$y^2 + 8y + 8$ , doubling the $4$ for both the middle term and the constant.

Fluency Practice

The parabola $x^2 = 16y$ has its vertex at the origin. Find the value of $p$ , the distance from the vertex to the focus.

What is the directrix of the parabola $x^2 = 16y$ ?

$y = 4$ , placing the directrix on the same side of the vertex as the focus.

$x = -4$ , treating the parabola as opening sideways because the equation begins with $x^2$ .

$y = -16$ , reading the directrix straight off the coefficient $16$ without solving $4p = 16$ .

$y = -4$ , since $4p = 16$ gives $p = 4$ and the directrix of an upward parabola is $y = -p$ .

A parabola has focus $(0, -5)$ and directrix $y = 5$ . Write its equation in the form $x^2 = 4py$ .

$Coordinate plane with a parabola opening left from the origin, a marked point at (-3, 0), and a dashed vertical line at x = 3$

In which direction does the parabola $y^2 = -12x$ open, and where is its focus?

Left, focus $(-3, 0)$ — the squared variable is $y$ , so the axis is horizontal, and $4p = -12$ gives $p = -3$ .

Right, focus $(3, 0)$ — the axis is horizontal, and the focus is taken $3$ units from the vertex on the positive side.

Down, focus $(0, -3)$ — the negative coefficient means the parabola opens downward from the origin.

Left, focus $(-12, 0)$ — the axis is horizontal and the coefficient $-12$ locates the focus directly.

A parabola has focus $(2, 0)$ and directrix $x = -2$ . Write its equation in the form $y^2 = 4px$ .

The parabola $y = 2x^2$ has vertex at the origin. What is its focal length $p$ ?

$p = 2$ , reading the focal length directly from the coefficient of $x^2$ .

$p = 8$ , computing $4a$ with $a = 2$ instead of dividing.

$p = \frac{1}{8}$ , since $a = \frac{1}{4p}$ means $p = \frac{1}{4a} = \frac{1}{4(2)}$ .

$p = \frac{1}{2}$ , taking the reciprocal of the coefficient $a = 2$ without the factor of $4$ .

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