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lozanna [386]
2 years ago
13

A jumping spider's movement is modeled by a parabola. The spider makes a single jump from the origin and reaches a maximum heigh

t 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)
Mathematics
1 answer:
aleksklad [387]2 years ago
4 0

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}

<u>(a) The equation</u>

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}

<u>(b) The focus, directrix and the axis of symmetry</u>

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}

Read more about parabolas at:

brainly.com/question/25237745

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