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1. Problem: Find the center and radius of (x - 2)^2 + (y + 3)^2 = 16. Then, graph the circle. Solution: Rewrite the equation in standard form. (x - 2)^2 + [y - (-3)]^2 = 4^2 The center is (2, -3) and the radius is 4. The graph is easy to draw, especially if you use a compass. The accompanying figure shows a graph of the solution.Back to top.
(x - h)^2 (y - k)^2 --------- + --------- = 1 a^2 b^2Example:
1. Problem: Graph x^2 + 16y^2 = 16. Solution: Multiply both sides by 1/16 to put the equation in standard form. x^2 y^2 --- + --- = 1 16 1 a = 4 and b = 1. The vertices are at (±4, 0) and (0, ±1). The points are on the axes because the equation tells us the center is at the origin, so the vertices have to be on the axes). Connect the vertices to form an oval, and you are done! See the accompanying figure to see a graph of the solution.Back to top.
x^2 y^2 --- - --- = 1 a^2 b^2Example:
1. Problem: Graph 9x^2 - 16y^2 = 144. Solution: First, multiply each side of the equation by 1/144 to put it in standard form. x^2 y^2 --- - --- = 1 16 9 We know that a = 4 and b = 3. The vertices are at (±4, 0). (Since we know the center is at the origin, we know that the vertices are on the x-axis.) The easiest way to graph a hyperbola is to draw a rectangle using the vertices and b, which is on the y-axis. Draw asymptotes through opposite corners of the rectangle. Next, draw the hyperbola. The accompanying figure is the graph of 9x^2 - 16y^2 = 144.Back to top.
1. Problem: Solve the following system of equations: x^2 + y^2 = 25 3x - 4y = 0 Solution: Graph both equations on the same coordinate plane. The points of intersection have to satisfy both equations, so be sure to check the solutions. (Both intersections do check.) This example figure shows the graphs and intersections (solutions) of both equations.Back to top.
Take the quiz on coordinate geometry. The quiz is very useful for either review or to see if you've really got the topic down.