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 16 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

Answer:

$\textsf{A)} \quad (x-40)^2=-100(y-16)$

B)  Focus = (40, -9)

    Directrix:  y = 41

    Axis of symmetry:  x = 40

Step-by-step explanation:

The x-intercepts of a parabola are the points at which the curve intercepts the x-axis (when y = 0).

The x-coordinate of the vertex of a parabola is halfway between the x-intercepts.

The y-coordinate of the vertex if the minimum or maximum height of the parabola.

Part A

A jumping spider's movement is modeled by a parabola.

Define the variables:

• x = horizontal distance of the spider
• y = height of the spider

From the information given:

• x-intercepts = (0, 0) and (80, 0)
• vertex = (40, 16)

Standard form of a parabola with a vertical axis of symmetry:

$(x-h)^2=4p(y-k) \quad \textsf{where}\:p\neq 0$

• Vertex:  (h, k)
• Focus:  (h, k+p)
• Directrix:  y = (k-p)
• Axis of symmetry:  x = h

If p > 0, the parabola opens upwards, and if p < 0, the parabola opens downwards.

Substitute the vertex (40, 16) and one of the x-intercept points (0, 0) into the formula and solve for p:

$\implies (0-40)^2=4p(0-16)$

$\implies 1600=-64p$

$\implies p=-25$

Substitute the vertex and the found value of p into the formula:

$\implies (x-40)^2=4(-25)(y-16)$

$\implies (x-40)^2=-100(y-16)$

Part B

Given:

• Vertex = (40, 16)  ⇒  h = 40 and k = 16
• p = -25

Substitute the given values into the formulas for focus, directrix and axis of symmetry:

Focus

⇒ (h, k+p)

⇒ (40, 16 + (-25)))

⇒ (40, -9)

Directrix

⇒ y = (k-p)

⇒ y = (16 - (-25))

⇒ y = 41

Axis of symmetry

⇒ x = h  

⇒ x = 40