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