If we have an ellipse with a center at (h,k), and radii of a and b, then the standard form for that ellipse will look like:
(x-h)^2/(a^2) + (y-k)^2/(b^2) = 1
So we have a radius along the x-axis and another along the y-axis. The larger radius will be along the major (longer) axis, and the smaller radius will be along the minor (shorter) axis. So the ellipse can be longer in the x-direction, or longer in the y-direction, depending on if a or b is larger.
The vertices are points where the major axis touches the ellipse, while the covertices are the points where the minor axis touches the ellipse. They are a and b units away from the center of the ellipse.
The foci are two points on the interior of the ellipse. The sum of the distance from any point on the ellipse to each focus is constant. The foci will be c units from the center of the ellipse (h,k) along the major axis. If a is the radius of the major axis and b is the radius of the minor axis (so major = x and minor = y),
c^2 = a^2 - b^2
If the major axis is the y-axis with radius b, and the minor axis is the x-axis with radius a,
c^2 = b^2 - a^2
Either way it is always major axis minus minor axis! So now we've talked about everything except domain range. These are trivial. The domain, being the set of all possible inputs/x-values for a function, will extend from the center of the ellipse minus a, to the center of the ellipse plus a. The range, being all possible y-values or outputs for a function, will extend from the center of the ellipse minus b to the center of the ellipse plus b.
Domain: [h-a, h+a]
Range: [k-b, k+b]
So now we know how to find everything we are looking for, given that our equation is already in standard form. So our task is to put the equation in standard form. To do this, we're going to have to factor both the x terms and y terms into perfect squares, and then make sure the right side of the equation is equal to one.
To factor into perfect squares, we're going to use a method called "complete the square." For complete the square, we need to make each quadratic (x and y) look like:
x^2 + bx + c
c can be zero, but if there's anything in front of the x^2 term, we need to factor it out.
4x^2 + 24x + y^2 + 2y = -33
We have a 4 in front of x^2. Let's factor 4 out of 4x^2 + 24x.
4(x^2 + 6x) + y^2 + 2y = -33
Great! Now we know that for the x terms, b = 6. We want to add (b/2)^2 to both sides of the equation, because this is what will "complete the square." So b/2 = 3, (b/2)^2 = 9. We have to add the same thing to both sides, but the 9 needs to be added inside the parentheses on the left hand side. But since the parentheses are being multiplied by 4, what we want to add to the right side is not 9 but 9*4=36.
So add 9 on the left inside, 36 on the right.
4(x^2 + 6x + 9) + y^2 + 2y = 3
Now inside the parentheses we have something that factors cleanly into a perfect square.
4(x + 3)^2 + y^2 + 2y = 3
Now to tackle the y terms. Nothing in front of y^2 to factor out. b=2. b/2 = 1. (b/2)^2 = 1. Add 1 to both sides.
4(x + 3)^2 + y^2 + 2y + 1 = 4
Now factor.
4(x + 3)^2 + (y + 1)^2 = 4
We're almost there. Got to make the right side equal to one. So divide both sides by 4.
(x + 3)^2 + (y + 1)^2/4 = 1
Looks like the radius along the x-axis is 1, and the radius along the y-axis is 2 (2^2=4). So the y-axis is the major axis and the x-axis is the minor axis. We can also see the center is at (-3,-1). So now we can find the vertices and covertices.
Vertices : ( -3, -3), ( -3, 1)
Covertices: ( -4, -1), ( -2, -1)
This tells us the domain and range too, since the domain will go from covertex to covertex and the range will go from vertex to vertex.
Domain: [ -4, -2]
Range: [ -3, 1]
What about the foci? Well we need to know c. Since y is the major axis,
c^2 = b^2 - a^2
c^2 = 2^2 - 1^2
c^2 = 3
c = sqrt(3)
So the foci will be on the y=axis, sqrt(3) units away from the center.
Foci: ( -3, -1 - sqrt(3) ), ( -3, -1 + sqrt(3) )
Plot the covertices and vertices, and draw the ellipse through them. Plot the center and the foci. We're done here!
