Find the vertex, focus, and directrix of the parabola.
x^2 - 2x + 8y + 9 = 0
x^2 - 2x +1 = -8y + 9+ 1
(x-1)^2 = -8(y-1.25)
vertex:(1,1.25)
focus:(1,-.75)
directrix: y=3.25
My teacher said that I have sign errors. I do not know where I went wrong.
the +9 when transferred to the other side should be -9
Why did you add 1 to each side?
To find the vertex, focus, and directrix of a parabola in the form ax^2 + bx + cy + d = 0, you can follow these steps:
1. Rewrite the equation in the form (x - h)^2 = 4p(y - k), where (h, k) is the vertex of the parabola and p is the distance from the vertex to the focus/directrix.
In your case, x^2 - 2x + 8y + 9 = 0, rearranging the terms gives us:
(x^2 - 2x) = -8(y + 1.125)
2. Complete the square for the x terms. Take half of the coefficient of x (-2) and square it (4) to add and subtract from the equation:
(x^2 - 2x + 1) = -8(y + 1.125) + 1
3. Simplify the equation:
(x - 1)^2 = -8(y + 1.125) + 1
Now the equation is in the desired form.
4. Identify the vertex from the equation. The vertex is at the point (h, k) = (1, -1.125).
5. Since the coefficient of y is negative, the parabola opens downward. Therefore, the focus of the parabola is located p units below the vertex, and the directrix is p units above the vertex.
6. Calculate the value of p by using the formula p = 1/(4|a|), where a is the coefficient of y. In this case, p = 1/(4 * |-8|) = 1/32.
7. Find the coordinates of the focus by adding p to the y-coordinate of the vertex. The focus is at (1, -1.125 + 1/32) = (1, -1.15625) approximately.
8. Finally, find the equation of the directrix. Since the directrix is p units above the vertex, the equation is y = -1.125 + 1/32 = -1.09375.
So, after rechecking the calculations, the corrected answers are:
Vertex: (1, -1.125)
Focus: (1, -1.15625)
Directrix: y = -1.09375
Please double-check your calculations and make sure to pay attention to the signs when completing the square.