Question

Help please! This time it's on a different question. Using a directrix of y = −2 and a focus of (2, 6), what quadratic function is created?

Answer 1

@Mehek14 @DogzCatz @Maria195

Answer 2

@tom982

Answer 3

The distance from a point on the parabola to the focus must be the same as the distance from that point to the line y=-2

Answer 4

http://hotmath.com/hotmath_help/topics/finding-the-equation-of-a-parabola-given-focus-and-directrix.html
- this will help as well

Answer 5

Let (x,y) be a point on the parabola, the distance squared from it to the point (2.6) is
\[
d^2= (x-2)^2 +(y-6)^2
\]

Answer 6

How much is the distance from (x,y) to the line y=-2?

Answer 7

\[
d^2=(y+2)^2
\]

Answer 8

Equate, Simplify and you get the answer

Answer 9

Umm, I'm not completely sure. Would it be f(x)= -1/8 (x-2)^2 -2?

Answer 10

...Am I right?

Answer 11

Or wait... Is it -1/16 (x-2)^2 -2? Now I'm not sure :/

Answer 12

Please help, I'm kind of confused now. :/

Answer 13

And I have another question. This one's a lot easier but my mind just blanked on me. Which polynomial identity will prove that 16 = 25 − 9?

Answer 14

Are you working it out?...

Answer 15

Nevermind, I just went ahead and answered because I couldn't wait any longer. I got it wrong for all of you who might use this as future reference :(

Answer 16

The equation of the parabola should be
\[
y=\frac{1}{16} \left(x^2-4 x+36\right)
\]

Answer 17

It is a parabola

Answer 18

It is good to be able to visualize these problems before working the equations.

You have a directrix of \(y=-2\) and a focus of \((2,6)\). The parabola passes between the directrix and the focus (it is the set of all points equidistant from directrix and focus), so with the focus above the directrix, we have a parabola which opens upward.

General form of a parabola which opens upward is \[y = a(x-h)^2+k\]where \((h,k)\) is the vertex of the parabola. It can also be written as \[y = \frac{1}{4p}(x-h)^2+k\]where \(p\)is half the distance between directrix and focus. If you have two points on the parabola, the former may be easier; if you have the focus and the directrix, the latter is easier.

Distance from focus to directrix is simply \[| 6-(-2)| = 8\]so \(p=8/2=4\)

That means our equation is \[y=\frac{1}{4(4)}(x-h)^2+k=\frac{1}{16}(x-h)^2+k\]

To find the vertex, remember that the vertex is the point which is closest to both the focus and the directrix, and thus the point halfway between them. Our vertex must be at \(x=2\) and \(y = -2 + p = -2+4 = 2\) or \((2,2)\). That gives us \(h = 2,\ k=2\) and our equation is
\[y = \frac{1}{16}(x-2)^2+2\]

Here's a graph showing the various points of interest:

Answer 19

And for your other question:

"Which polynomial identity will prove that 16 = 25 − 9"

\[\sqrt{25} = 5\]\[\sqrt{9}=3\]You can express\[25-9\]as the difference of two squares:
\[(x-a)(x+a)= x*x+a*x-a*x-a*2 = x^2-a^2\]

if we set \[x=5\]and \[a=3\]we have \[(5+3)(5-3) = 5^2-3*5+3*5-3*3 = 5^2-3^2\]but we also know that \[(5+3)(5-3) = 8*2 = 16\]therefore\[5^2-3^2 = 16\]even if we couldn't calculate the values of \(5^2\) and \(3^2\) for some bizarre reason.

@Kimberly_PR