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)

The equation of the parabola is [tex]\mathbf{y = -\frac{1}{640}(x - 80)^2 + 10}[/tex]
The focus is [tex]\mathbf{Focus = (80, -1760)}[/tex]
The directrix is [tex]\mathbf{y = \frac{1}{640}}[/tex]
The axis of the symmetry of parabola is: [tex]\mathbf{x = 80}[/tex]

From the question, we have:

[tex]\mathbf{Vertex: (h,k) = (80,10)}[/tex]

[tex]\mathbf{Origin: (x,y) = (0,0)}[/tex]

The equation of a parabola is:

[tex]\mathbf{y = a(x - h)^2 + k}[/tex]

Substitute the values of origin and vertex in [tex]\mathbf{y = a(x - h)^2 + k}[/tex]

[tex]\mathbf{0 = a(0 - 80)^2 + 10}[/tex]

[tex]\mathbf{0 = a(- 80)^2 + 10}[/tex]

[tex]\mathbf{0 = 6400a + 10}[/tex]

Collect like terms

[tex]\mathbf{6400a =- 10}[/tex]

Solve for a

[tex]\mathbf{a =- \frac{1}{640}}[/tex]

Substitute the values of a and the vertex in [tex]\mathbf{y = a(x - h)^2 + k}[/tex]

[tex]\mathbf{y = -\frac{1}{640}(x - 80)^2 + 10}[/tex]

The focus of a parabola is:

[tex]\mathbf{Focus = (h, \frac{k+1}{4a})}[/tex]

Substitute the values of a and the vertex in [tex]\mathbf{Focus = (h, \frac{k+1}{4a})}[/tex]

[tex]\mathbf{Focus = (80, \frac{10+1}{4 \times -\frac{1}{640}})}[/tex]

[tex]\mathbf{Focus = (80, -\frac{11}{\frac{1}{160}})}[/tex]

[tex]\mathbf{Focus = (80, -11\times 160)}[/tex]

[tex]\mathbf{Focus = (80, -1760)}[/tex]

The equation of the directrix is:

[tex]\mathbf{y = -a}[/tex]

So, we have:

[tex]\mathbf{y = \frac{1}{640}}[/tex] ----- the directrix

The axis of symmetry is:

[tex]\mathbf{x = -\frac{b}{2a}}[/tex]

We have:

[tex]\mathbf{y = -\frac{1}{640}(x - 80)^2 + 10}[/tex]

Expand

[tex]\mathbf{y = -\frac{1}{640}(x^2 -160x + 6400) +10}[/tex]

Expand

[tex]\mathbf{y = -\frac{1}{640}x^2 +\frac{1}{4}x - 10 +10}[/tex]

[tex]\mathbf{y = -\frac{1}{640}x^2 +\frac{1}{4}x }[/tex]

A quadratic function is represented as:

[tex]\mathbf{y = ax^2 + bx + c}[/tex]

So, we have:

[tex]\mathbf{a = -\frac{1}{640}}[/tex]

[tex]\mathbf{b = \frac{1}{4}}[/tex]

Recall that:

[tex]\mathbf{x = -\frac{b}{2a}}[/tex]

So, we have:

[tex]\mathbf{x = -\frac{1/4}{2 \times -1/640}}[/tex]

[tex]\mathbf{x = \frac{1/4}{1/320}}[/tex]

This gives

[tex]\mathbf{x = \frac{320}{4}}[/tex]

[tex]\mathbf{x = 80}[/tex]

Hence, the axis of the symmetry of parabola is: [tex]\mathbf{x = 80}[/tex]