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Tyrese Nolan
Answered 2020-03-16 14:26:56

This equation is equal to the first one because it produces the same results, always. ... TL;DR - The circle equation is what you get when you multiply all terms from the ellipse equation by the radius. x^2/a^2 + y^2/b^2 = 1 is an ellipse equation. Well, a circle has a radius where a and b are the same.

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This ellipse is centered at the origin and has a horizontal axis of length 26 and a vertical axis of length 12 What is its equation?





By taking a coordinate system with origin at the center of the ellipse, and x-axis along the major axis, and y-axis along the minor axis, then the ellipse intercepts the x-axis at -5 and 5, and the y-axis at -2 and 2. So that the equation of the ellipse x2/a2 + y2/b2 = 1 becomes x2/52 + y2/22 = 1 or x2/25 + y2/4 = 1.


Ellipse formula, centered at the origin, where the vertical axis is the major axis: x2/b2 + y2/a2 = 1, a > b Since the major axis is 8, then a = 4. Since the minor axis is 4, then b = 2. Thus, the equation of the ellipse is: x2/4 + y2/16 = 1.


The equation is based on formula (x - h)square / A square + (y-k)square / B square = 1. To apply to the above ellipse the equation would be similar to (x- 0) square/ 14 square + (2014 - 0) square / 16 square.



x2/a2 + y2/b2 = 1, is the equation of an ellipse with semi-major axes a and b (that's the equivalent of the radius, along the two different axes), centered in the origin.


You know the formula for the area of a circle of radius R. It is Pi*R2. But what about the formula for the area of an ellipse of semi-major axis of length A and semi-minor axis of length B? (These semi-major axes are half the lengths of, respectively, the largest and smallest diameters of the ellipse--- see Figure 1.) For example, the following is a standard equation for such an ellipse centered at the origin: (x2/A2) + (y2/B2) = 1. The area of such an ellipse is Area = Pi * A * B , a very natural generalization of the formula for a circle!


The equation of any circle, centered at the origin, is x2 + y2 = r2, where r is the radius.


The tricky part is making an ellipse at a point different from the origin.


The general equation of a circle is given by the formula(x - h)2 + (x - k)2 = r2, where (h, k) is the center of the circle, and r its radius.Since the center of the circle is (0, 0), the equation reduces tox2 + y2 = r2So that the equation of our circle is x2 + y2 = 36.






If (3, -6) is on a circle with its center at the origin, the equation for the circle is: x2 + y2 = (sqrt)45


It seems that it is the equation of an ellipse, x2/b2 + y2/a2 = 1, with center at the origin, a vertical major axis with length 2a, and a horizontal minor axis with length 2b. x2/36 + y2/64 = 1 Since you are looking for the x-intercepts, they are (√b2, 0) = (√36, 0) = (6, 0) and (-√b2, 0) = (-√36, 0) = (-6, 0)


The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.The equation of a circle centered at the origin is x2 + y2 = r2; in this case, x2 + y2 = 64.




It is the equation of a circle with radius of 6 and its center at the origin.


The standard form of the equation of a hyperbola with center at the origin isx2/a2 - y2/b2 = 1 where the transverse axis lies on the x-axis,ory2/a2 - x2/b2 = 1 where the transverse axis lies on the y-axis.The vertices are a units from the center and the foci are c units from the center.For both equations, b2 = c2 - a2. Equivalently, c2 = a2 + b2.Since we know the length of the transverse axis (the distance between the vertices), we can find the value of a (because the center, the origin, lies midway between the vertices and foci).Suppose that the transverse axis of our hyperbola lies on the x-axis.Then, |a| = 24/2 = 12So the equation becomes x2/144 - y2/b2 = 1.To find b we need to know what c is.


x2 + y2 = r2, the equation of a circle centered at the origin. If you want to make the circle larger, increase the radius length.



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