There's an excellent formula that I concocted 5 days ago to always give the equation of a parabola when only the directrix and focus is given. Here is the equation:
Let a = x-coordinate of focus
Let b = y-coordinate of focus
Let k= equation of line of the directrix
\(y=\frac{(x-a)^2}{2(b-k)}+\frac{b+k}{2}\)
All you have to do is plug into this formula, simplify, and you're done. Let's try this together:
a=-2
b=4
k=6
\(y=\frac{(x-a)^2}{2(b-k)}+\frac{b+k}{2}\) | Plug in the appropriate values that are given by the focus and directrix. |
\(y=\frac{(x-(-2))^2}{2(4-6)}+\frac{4+6}{2}\) | Let's clean this up a bit, shall we? |
\(y=\frac{(x+2)^2}{-4}+5\) | Technicaly, you could stop here and call it a day, but I am going to attempt to make it look even cleaner! I'll expand the \((x+2)^2\) |
\(y=-\frac{x^2+4x+4}{4}+5\) | After expanding it, I'll change 5 into an improper fraction that I can add to the current fraction. |
\(y=-\frac{x^2+4x+4}{4}+\frac{20}{4}\) | Since the fractions have common denominators. Add the fractions together, but you have to be very attentive to how you do this. Notice how there is a negative. I'm going to get rid of this because you can't combine a negative fraction with a positive one; it just doesn't work. |
\(y=\frac{-(x^2+4x+4)}{4}+\frac{20}{4}\) | I'm distributing that negative because you cannot combine a negative and positive fraction. |
\(y=\frac{-x^2-4x-4}{4}+\frac{20}{4}\) | Now, you can add the fractions together. Now you add the fractions together normally. |
\(y=\frac{-x^2-4x+16}{4}\) | Break off the two last terms from the current fraction. You'll see why. |
\(y=\frac{-x^2}{4}+\frac{-4x+16}{4}\) | The rightmost fraction can be simplified because a factor of 4 goes into both terms. Wow! |
\(y=\frac{-x^2}{4}-x+4\) | |
\(y=-\frac{1}{4}x^2-x+4\) | This is your final equation. |
No, I did not just pull that formula out of the air! I derived it myself. If you are wondering why this formula will work 100% of the time, I explained it, in detail, here where someone asked a similar question to yours.
To figure out this problem, we must know a few formulas that relate to a circle. Of course, you must know the area formula for a circle:
Let A= Area of a circle
Let r= radius of the circle:
\(A_{\circ}=\pi r^2\)
However, there is a problem currently, because we do not know the circumference of this circle! However, we can solve for the radius if we know the circumference. The formula for finding the circumference is:
Let C = circumference of a circle
Let r = radius
\(C_{\circ}=2\pi r\)
We know the circumference of the circle because it is given; the circumference is 9.42 inches. Let's solve for r:
\(C_{\circ}=2\pi r\) | This is the formula for finding the circumference of a circle. We already know the circumference, so plug it into this equation |
\(9.42=2\pi r\) | Divide by 2 on both sides of the equation |
\(4.71=\pi r\) | Divide by \(\pi\) on both sides. |
\(\frac{4.71}{\pi}=r\) | I am going to leave the r in this form because I want my final answer to be as exact as possible. |
Now that we know what r equals, we can substitute this into the area formula for a circle:
\(A_{\circ}=\pi r^2\) | Substitute \(r\hspace{1mm}\text{for}\hspace{1mm}\frac{4.71}{\pi}\) |
\(A_{\circ}=\pi (\frac{4.71}{\pi})^2\) | We could simply input this into the calculator, but we can actually simplify this further. Let's do it! First, do \((\frac{4.71}{\pi})^2\). Remember that the exponent is distributed to both the numerator and denominator |
\(A_{\circ}=\frac{\pi}{1}*\frac{22.1841}{\pi^2}\) | Multiply the fractions together. |
\(A_{\circ}=\frac{22.1841\pi}{\pi^2}\) | Now, we will utilize a fraction rule stating that \(\frac{a^b}{a^c}=a^{b-c}\) |
\(A_{\circ}=22.1841\pi^{-1}\) | Now, we will use a power rule that says that \(a^b=\frac{1}{a^b}\hspace{3mm},b<0\) |
\(A_{\circ}=\frac{22.1841}{\pi}\) | You cannot simplify further, so evaluate with a calculator now. |
\(A_{\circ}=\frac{22.1841}{\pi}\approx7.0614in^2\) | Of course, keep units in your answer. |