SOLUTION: Write the equation for the quadratic in standard form. With the focus at (6,2), the vertex at (6,-2) and the directrix line at y=-6. I know that the answer is y=x^2/10-6x/5+2/5

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Write the equation for the quadratic in standard form. With the focus at (6,2), the vertex at (6,-2) and the directrix line at y=-6. I know that the answer is y=x^2/10-6x/5+2/5       Log On


   



Question 1056473: Write the equation for the quadratic in standard form. With the focus at (6,2), the vertex at (6,-2) and the directrix line at y=-6.
I know that the answer is y=x^2/10-6x/5+2/5 but I don't have any idea how to reach this solution!! PLEASE HELP MEEE

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Something is wrong with what you have stated above.
Either
y=x%5E2%2F10-6x%2F5%2B2%2F5
is not the answer or else the vertex is not (6,-2)
The vertex of any parabola is a point on the parabola
and therefore must satisfy the equation of the parabola.
However, (6,-2) DOES NOT satisfy the equation!
y=x%5E2%2F10-6x%2F5%2B2%2F5
Proof that it doesn't:
%28-2%29=%286%29%5E2%2F10-6%286%29%2F5%2B2%2F5
-2=36%2F10-36%2F5%2B2%2F5
-2=18%2F5-36%2F5%2B2%2F5
-2=-16%2F5
That is clearly false.
The parabola that you say is the answer
has vertex (6,-16/5),
focus (6,-7/10),
and directrix y = -57/10.
------------------------
The parabola that has focus at (6,2), vertex at (6,-2)
the directrix line at y=-6, has this equation:
y+=+x%5E2%2F16-3x%2F4%2B1%2F4
I would have shown you how to get that, but since the
answer you said is correct is incorrect, I thought
I'd better wait until I was sure you had everything
correct before doing it. If you want to discuss it
with me, you may do so in the thank-you note form
below and I'll get back to you by email.
Edwin