Graph the ellipse and locate the foci.
16x^2 +9y^2= 144
if a>b,
x^2/a^2 + y^2/b^2 = 1
has foci at (±c,0) where a^2=b^2+c^2
To graph the ellipse and locate the foci of the equation 16x^2 + 9y^2 = 144, we can follow these steps:
Step 1: Rewrite the equation in standard form
To find the standard form of the ellipse, we divide both sides of the equation by 144 to make the right side of the equation equal to 1:
(x^2)/(9) + (y^2)/(16) = 1
Step 2: Identify the values of a and b
The standard form of the ellipse equation is (x^2)/(a^2) + (y^2)/(b^2) = 1. By comparing this equation with our given equation, we can see that a^2 = 9 and b^2 = 16.
Step 3: Find the center of the ellipse
The center of the ellipse is always at the point (h, k) in the standard form equation. In this case, (h, k) = (0, 0).
Step 4: Calculate the values of a and b
Based on the values of a^2 and b^2 we found in step 2, a = 3 and b = 4.
Step 5: Find the foci
The foci of an ellipse are located on the major axis at points (h, k + c) and (h, k - c), where c = sqrt(a^2 - b^2).
Plugging in the values of a and b, we have c = sqrt(3^2 - 4^2) = √(9 - 16) = √(-7), which is an imaginary number. Since c is imaginary, the given ellipse has no real foci.
Step 6: Graph the ellipse
Now that we have all the necessary information, we can graph the ellipse. Using the values we found, the semi-major axis (a) is 3, and the semi-minor axis (b) is 4.
Plot the center at (0, 0), then plot points on the x and y axes that are ±a and ±b units away from the center. In this case, we plot the points (-3, 0), (3, 0), (0, -4), and (0, 4) to represent the ends of the major and minor axes.
Using these points, sketch a smooth curve connecting them. The resulting graph will be an ellipse centered at (0, 0) with a horizontal major axis of length 6 (from -3 to 3) and a vertical minor axis of length 8 (from -4 to 4).
Therefore, the graph of the ellipse is centered at the origin with a major axis of length 6 and a minor axis of length 8, but it has no real foci.