Graph 16x^2 + 9y^2 = 144
I know that you cannot graph this but the graph of this equation should be a circle I think.
I have points (3,0), (0,4), (-3, 0), and (0,-4) graphed. Is there any more points that I could use to graph?
Thanks.
nope, ellipse
4^2 x^2 + 3^2 y ^2 = 12^2
(4^2/12^2) x^2 + (3^2/12^2)y^2 = 1
x^2 /3^2 + y^2/ 4^2 = 1
form is
(x-h)^2/a^2 + (y-k)^2/b^2 = 1 for ellipse centered at (h,k)
so
ellipse with center at (0,0)
horizontal half length = a = 3
vertical half height = b = 4
You can put in all the intermediate points you wish, like x = 1 and x = -1 (same y answers)
and x = 2 and x = -2 (again same y answers).
This equation graphs as an ellipse, not a circle. An ellipse is a kind of flattended circle.
The equation can be rewritten, after dividing both sides by 144 and recognizing some perfect squares,
(x/3)^2 + (y/4)^2 = 1
The lengths of the major and minor axes are 3x2 = 6 and 4x2 = 8 ,(twice the numbers in the denominator). The center of the ellipse is the origin, (0,0).
I suggest you plot a few more points to more accurately graph the ellipse. For example, if x = + or -1,
y = +/- 3.77
major axis 20 units long and parallel to y-axis minor axis 6 units long center at (4,2)
y=5/x
You are correct that the equation 16x^2 + 9y^2 = 144 represents a circle. To graph this equation, we can follow a few steps:
Step 1: Rearrange the equation in standard form for a circle, which is (x - h)^2 + (y - k)^2 = r^2. Here, h and k represent the coordinates of the center of the circle, and r represents the radius.
16x^2 + 9y^2 = 144
Divide both sides of the equation by 144 to put it in standard form:
x^2/9 + y^2/16 = 1
Step 2: Compare the equation with the standard form. We can see that the equation has a coefficient of 1 for both x^2 and y^2. This means that the circle is centered at the origin (0, 0).
Step 3: Find the radius of the circle. In the standard form, the radius squared is the value on the right side of the equation. In this case, the radius squared is 1. Therefore, the radius of the circle is 1.
Now, let's find additional points to graph the circle. The points you have already graphed are correct and lie on the circle. However, it is always helpful to have more points to get a more accurate representation of the shape of the circle.
To find additional points, you can substitute different x or y values into the equation and solve for the corresponding y or x values.
For example, let's find a point when x = 2:
x^2/9 + y^2/16 = 1
(2^2)/9 + y^2/16 = 1
4/9 + y^2/16 = 1
y^2/16 = 1 - 4/9
y^2/16 = 5/9
y^2 = (5/9) * 16
y^2 = 80/9
y = ±√(80/9)
Therefore, when x = 2, we have two points that can be used to graph the circle: (2, √(80/9)) and (2, -√(80/9)).
By substituting different x values and finding the corresponding y values, you can obtain multiple points to plot on the graph and get a clearer picture of the circle. Repeat this process for different x values if more points are needed.
Remember, a circle is symmetric, so if you find one point on the circle, there will be another point on the opposite side of the circle with the same y-coordinate.
I hope this helps you graph the equation of the circle!