The graph of 11x - x^2 - 4y^2 = 2y - 16 is the graph of a(n) ________.
a. circle
b. ellipse
c. hyperbola
d. parabola
C?
if the signs of x^2 and y^2 terms are the same it is a circle or ellipse
If the coefficents are different, it's an ellipse
Since you have
11x - x^2 - 4y^2 = 2y - 16
x^2 + 4y^2 - 11x + 2y = 16
you have an ellipse. You will wind up with something of the form
(x-h)^2/4 + (y-k)^2 = 1
Thank you!
To determine the type of graph represented by the equation 11x - x^2 - 4y^2 = 2y - 16, we can rearrange it to a standard form that reveals the characteristics of the graph.
First, let's move all the terms to one side of the equation:
11x - x^2 - 4y^2 - 2y + 16 = 0
Next, let's group the x and y terms separately:
(-x^2 + 11x) + (-4y^2 - 2y) + 16 = 0
Now, let's complete the square to express the equation in a useful form. Starting with the x terms:
-1(x^2 - 11x) + (-4y^2 - 2y) + 16 = 0
To complete the square for the x terms, we need to take half the coefficient of x (-11/2), square it (121/4), and add it to both sides of the equation:
-1(x^2 - 11x + 121/4) + (-4y^2 - 2y) + 16 + 121/4 = 0
Similarly, for the y terms, we take half the coefficient of y (-2/2) and square it (1) to complete the square for the y terms:
-1(x^2 - 11x + 121/4) - 4(y^2 + 2y + 1) + 16 + 121/4 - 4 = 0
Now, let's simplify the equation further:
-1(x - 11/2)^2 - 4(y + 1)^2 + 169/4 - 16 - 4 = 0
Combining like terms, we get:
-1(x - 11/2)^2 - 4(y + 1)^2 + 25/4 = 0
Now, we can see that the equation has a negative coefficient in front of the x term and a negative coefficient in front of the y term. This tells us that the graph represents an ellipse. Therefore, the correct answer is b. ellipse.
So, the graph of 11x - x^2 - 4y^2 = 2y - 16 is the graph of an ellipse.