Answer:
\((x-6)^{2}=20(y+5)^{2}\)
Step-by-step explanation:
The standard for for the equation of a parabola is \((x-h)^{2}=4p(y-k)\)
The focus is given as \((h, k+p)\)
The directrix is given by \(y=k-p\)
- Comparing focus given as \((6,0)\) to formula of focus \((h,k+p)\) , we see that \(h=6\) and \(k+p=0\)
- Comparing directrix given as \(y=-10\) to formula of directrix \(y=k-p\) , we can write \(k-p=-10\)
We can use the two system of equations [ \(k+p=0\) and \(k-p=-10\) ] to solve for \(k\) and \(p\).
Adding the two equations gives us:
\(2k=-10\\k=\frac{-10}{2}\\k=-5\)
Using this value of \(k\) , we can find \(p\) by plugging this value in either equation. Let's put it in Equation 1. We have:
\(k-p=-10\\-5-p=-10\\-5+10=p\\p=5\)
Now, that we know \(h=6\) , \(k=-5\) , and \(p=5\) , we can plug these values in the standard form equation to figure out the parabola's equation. Doing so and rearranging gives us:
\((x-6)^{2}=4(5)(y-(-5))^{2}\\(x-6)^{2}=20(y+5)^{2}\\\)
This is the equation of the parabola with focus \((6,0)\) and directrix \(y=-10\)
