## Real World Applications – Algebra I

### Topic

Designing a Mirror for a Telescope in Space

### Student Exploration

Astronomers use telescopes to get a good view to see what’s out there in space. For this activity, we’re going to start to look at the importance of designing a mirror for one of these telescopes and represent its design using quadratics.

In this particular activity, we’re given that the circular area of this telescope is 14.12 square meters. This means that the radius is 2.12 meters, and the diameter is 4.24 meters.

We also know that there is a particular equation that represents the focal length of the parabola of the mirror. This equation is \begin{align*}F = \frac{1}{(4a)}\end{align*}

\begin{align*}F = \frac{1}{(4a)}\end{align*}

\begin{align*}8 = \frac{1}{(4a)}\end{align*}

\begin{align*}8(4a) &= 1\\
32a &= 1\\
a &= \frac{1}{32} = 0.03125 \ meters\end{align*}

Now that we have our \begin{align*}x-\end{align*}

The vertex form of a quadratic is \begin{align*}y = a(x -h)^2 + k\end{align*}

\begin{align*}y = 0.03125(x - 2.12)(x - 2.12)\end{align*}

\begin{align*}y = 0.03125(x^2 - 2.12x - 2.12x + 4.4944)\end{align*}

\begin{align*}y = 0.03125(x^2 - 4.24x + 4.4944)\end{align*}

\begin{align*}y = 0.03125x^2 - 0.1325x + 0.14045\end{align*}

Now we have our quadratic equation in standard form, meaning that on the right side of the equation, we look at the higher order of the terms. In this case, \begin{align*}x^2\end{align*}

Also, when looking at this equation, we see that the \begin{align*}y-\end{align*}

If we were to graph this quadratic equation, we will have the following:

The \begin{align*}x-\end{align*}

Now, just by looking at this graph, it doesn’t have the shape of a symmetrical mirror for an astronomer. Why not?

We know that the radius of the mirror is 2.12 meters, and the diameter is 4.24 meters. Let’s change our graph so that our graph only reaches a maximum width, or horizontal distance, of 4.24 meters.

Much better! What observations can you make about this graph and the mirror of the telescope?

Let’s dig a little bit deeper into the Algebra of this quadratic equation and analyze what the discriminant means.

For the equation, the discriminant is the part under the square root sign when finding the \begin{align*}x-\end{align*}

When we use the quadratic formula to find the \begin{align*}x-\end{align*}

\begin{align*}a &= 0.03125, b = -0.1325, c = 0.14045\\
x &= -(-0.1325) +- \frac{\sqrt{((-0.1325)^2 - 4(0.03125)(0.14045))}}{2(0.03125)}\\
x &= 0.1325 +- \frac{\sqrt{(0.01756 - 0.01756)}}{0.0625}\end{align*}

Let’s look at the discriminant, which is the part under the square root. We see that so far, the discriminant is “\begin{align*}0.01756 - 0.01756\end{align*}

After reviewing this concept, we know that if the discriminant is zero, then there is only one \begin{align*}x-\end{align*}

### Resources Cited

http://www.thefutureschannel.com/pdf/algebra/telescope_2.pdf