complete the square to identify what type of conic you have, identify the key parts indicated and then graph conic.
parabola: vertex,focus, directrix, focal diameter.
ellipse: center,vertices, foci, eccentricity.
hyperbola: center, vertices, foci, slope of asymptotes.
9x^2+36=y^2+36x+6y
9x^2+36=y^2+36x+6y
9x^2-36x -y^2-6y = -36
9(x^2-4x) - (y^2+6y) = -36
9(x^2-4x+4) - (y^2+6y+9) = -36 + 9*4 - 9
9(x-2)^2 - (y+3)^2 = -9
(y+3)^2 - 9(x-2)^2 = 9
(y+3)^2/9 - (x-2)^2 = 1
That's a standard form, so you should be able to provide its description now.
To complete the square and identify the type of conic, we need to rewrite the equation in a standard form. Let's work through it step by step:
1. Rearrange the equation by grouping the x-terms and y-terms:
9x^2 - 36x + 36 = y^2 + 6y - 36
2. Complete the square for both the x-terms and y-terms separately.
For the x-terms:
- Take half of the coefficient of x (-36) and square it: (-36/2)^2 = 324.
Add that value to both sides of the equation:
9x^2 - 36x + 324 = y^2 + 6y - 36 + 324
Simplify:
9(x^2 - 4x + 36) = y^2 + 6y + 288
For the y-terms:
- Take half of the coefficient of y (6) and square it: (6/2)^2 = 9.
Add that value to both sides of the equation:
9(x^2 - 4x + 36) + 9 = y^2 + 6y + 9 + 288
Simplify:
9(x^2 - 4x + 36) + 9 = y^2 + 6y + 297
3. Simplify each side:
9(x^2 - 4x + 36) + 9 = y^2 + 6y + 297
9x^2 - 36x + 324 + 9 = y^2 + 6y + 297
9x^2 - 36x + 333 = y^2 + 6y + 297
4. Now, complete the square for both the x-terms and y-terms by adding constant terms to each side:
9(x^2 - 4x + 4) + 333 = y^2 + 6y + 9 + 297 + 4 * 9
9(x^2 - 4x + 4) + 333 = y^2 + 6y + 9 + 297 + 36
Simplify:
9(x - 2)^2 + 333 = (y + 3)^2 + 342
5. Rewrite the equation in standard form:
9(x - 2)^2 - (y + 3)^2 = 9
Now we can identify the type of conic:
From the equation, we can see that the powers of x and y are both squared, which suggests that this is either an ellipse or a hyperbola.
To determine which conic it is, we can check the coefficients:
- If the coefficients of both squared terms have the same sign, we have an ellipse.
- If the coefficients of the squared terms have opposite signs, we have a hyperbola.
In this case, we have 9 for the x-squared term and -1 for the y-squared term. Since the coefficients have opposite signs, we have a hyperbola.
Now, let's identify the key parts and graph the hyperbola:
Center: The center of the hyperbola is (h, k), which can be found by moving the constants to the other side of the equation. In this case, (h, k) = (2, -3).
Vertices: The vertices of a hyperbola are given by (h ± a, k), where a is the distance from the center to the vertices. From the equation, we have a^2 = 1, so a = 1. Therefore, the vertices are (1, -3) and (3, -3).
Foci: The foci of a hyperbola are given by (h ± c, k), where c is the distance from the center to the foci. From the equation, we have c^2 = a^2 + b^2 = 1 - 9 = -8. Since we have a negative value, there are no real foci.
Slope of Asymptotes: The slope of the asymptotes of a hyperbola can be found using the equation y - k = ±(a/b)(x - h). In this case, a = 1, and b = sqrt(|c^2 - a^2|) = sqrt(|-8 - 1|) = sqrt(9) = 3. Thus, the slope of the asymptotes is ±(1/3).
Now that we have identified all the key parts, we can graph the hyperbola on a coordinate plane with the center, vertices, and asymptotes.
Remember, the equation for the hyperbola is:
9(x - 2)^2 - (y + 3)^2 = 9