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# Equation of Circle, Parabola & Ellipse

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Find the equations of the following:

a) A circle where (4,-1) and (-6,6) are endpoints of a diameter.
b) A parabola with a focus at (3,4) and a directrix at x=-1.
c) An ellipse with vertices at (-2,5) and (-2,1) and a minor axis 2 units long.

https://brainmass.com/math/geometry-and-topology/equation-circle-parabola-ellipse-405538

## SOLUTION This solution is FREE courtesy of BrainMass!

Find the equations of the following:
a) A circle where (4,-1) and (-6, 6) are endpoints of a diameter.
We know that the centre of the circle is the mid-point of the diameter.
Therefore, mid-point of the diameter =
= (-1, 5/2)
Hence centre of the circle = (-1, 5/2)
Radius of the circle = the distance between the centre and one end point
=
=
The standard equation of a circle with centre (h, k) and radius r is given by,
(x - h)2 + (y - k)2 = r2
Therefore, equation of the circle is given by,

That is, x2 + 2x + 1 + y2 - 5y + 25/4 = 149/4
Simplifying, we get,
x2 + y2 + 2x - 5y - 30 = 0, which is the required, equation.
b) A parabola with a focus at (3, 4) and a directrix at x=-1.
Let (x, y) be any point on the parabola. The point to the right of the point (x, y) and lying on the directrix is given by (-1, y).
We know that the distance from the point (x, y) to the focus and to the directrix are the same.
That is, distance between (x, y) and (3, 4) = distance between (x, y) and (-1, y)
That is,
Squaring both sides, we get,

That is, (y - 4)2 = x2 + 2x + 1 - x2 + 6x - 9
That is, (y - 4)2 = 8x - 8
That is, (y - 4)2 = 8(x - 1)
c) An ellipse with vertices at (-2, 5) and (-2, 1) and a minor axis 2 units long.
Clearly the y-axis is the major axis.
The standard form of an ellipse with major axis of Y is
, where (h, k) is the centre of the ellipse. The two points given are the end points of the major axis and the midpoint will be the center of the ellipse.
Using the mid-point formula we can find the centre as follows.
= (-2, 3)
Since the length of minor axis is 2, b = 2/2 = 1
Now, a = 5 - 3 = 2
Hence the required equation of the ellipse is given by,

That is,
Simplifying, we get,
4x2 + y2 + 16x - 6y +21 = 0

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