Question

A jumping spider's movement is modeled by a parabola. The spider makes a single jump from the origin and reaches a maximum height of 10 mm halfway across a horizontal distance of 80 mm.

Part A: Write the equation of the parabola in standard form that models the spider's jump. Show your work. (4 points)

Part B: Identify the focus, directrix, and axis of symmetry of the parabola. (6 points)

Answer 1

The spider's movement is an illustration of a parabola.

• The equation of a parabola is: $\mathbf{y = -\frac{1}{160}(x - 40) + 10}$
• The focus of a parabola is: (40,-30)
• The axis of symmetry is: $\mathbf{x = 40}$
• The directrix is:$\mathbf{y = 50}$

(a) The equation

The spider passes through the origin.

So, we have:

$\mathbf{(x,y) = (0,0)}$

The spider jumps to a maximum height of 10mm, midway 80mm.

So, the vertex is:

$\mathbf{(h,k) = (40,10)}$

The equation of a parabola is:

$\mathbf{y = a(x - h)^2 + k}$

So, we have:

$\mathbf{0 = a(0 - 40)^2 + 10}$

Subtract 10 from both sides

$\mathbf{a(0 - 40)^2 =- 10}$

$\mathbf{1600a =- 10}$

Solve for a

$\mathbf{a =- \frac{1}{160}}$

Substitute $\mathbf{a =- \frac{1}{160}}$ and $\mathbf{(h,k) = (40,10)}$ in $\mathbf{(h,k) = (40,10)}$

$\mathbf{y = -\frac{1}{160}(x - 40) + 10}$

(b) The focus, directrix and the axis of symmetry

The focus of a parabola is:

$\mathbf{Focus= (h, k + p)}$

Where:

$\mathbf{p = \frac{1}{4a}}$

So, we have:

$\mathbf{p = \frac{1}{4 \times -1/160}}$

$\mathbf{p = -\frac{160}{4}}$

$\mathbf{p = -40}$

So, we have:

$\mathbf{Focus = (40,10-40)}$

$\mathbf{Focus = (40,-30)}$

The axis of symmetry is:

$\mathbf{x = h}$

So, we have:

$\mathbf{x = 40}$

The directrix is:

$\mathbf{y = k - p}$

$\mathbf{y = 10 + 40}$

$\mathbf{y = 50}$

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