Write your answer in the form Ax^2+Cy^2+Dx+Ey+F=0.
A parabola with vertex (2,-1) opens right with a focal length "p" of 3.
I don't understand, please help!
Sure, I'll be happy to help you!
To write the equation of a parabola in the desired form Ax^2+Cy^2+Dx+Ey+F=0, we need to gather some information.
Given that the vertex of the parabola is (2, -1) and it opens to the right, we can determine that the general equation of the parabola is of the form (y - k)^2 = 4p(x - h), where (h, k) represents the vertex of the parabola and p represents the focal length.
In our case, the vertex is (2, -1), so the equation becomes (y + 1)^2 = 4p(x - 2).
Furthermore, we are given that the focal length p is 3. Substituting this value into the equation, we have (y + 1)^2 = 4(3)(x - 2).
Expanding the equation, we get (y + 1)^2 = 12(x - 2). Simplifying further gives (y + 1)^2 = 12x - 24.
Finally, rearranging the equation in the desired form, we have 0 = -12x + (y + 1)^2 + 24.
Therefore, the equation of the parabola is -12x + (y + 1)^2 + 24 = 0.
I hope this explanation helps you understand how to write the equation of a parabola in the given form! If you have any further questions, feel free to ask.