Question

What is the equation of the quadratic graph with a focus of (6, 0) and a directrix of y = −10 ?

Answer 1

(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 2

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  $y=-10$