x^2+y^2=100 and (X^2)/16-((y-2)^2)/4=1?

Question

oops i asked this earlier but it was interpreted wrong.
the first one reads, x squared plus y squared equals 100.
the second one reads x squared divided by 16, minus (y-2)^2 divided by four, equals 1.
How do you solve this system of equations, and be able to graph it?...

I know that you have to eventually subtract x^2-4(y-2)^2=16 from the first equation, but how do you solve the -4(y-2)^2 part of it, so as to get a y and y^2 value?

Answer 1

I don't quite agree with James.
(X^2)/16 + ((y-2)^2)/4=1 is an ellipse, but we have

(X^2)/16-((y-2)^2)/4=1, which is a hyperbola. Its asymptotes are

x/4 + (y-2)/2 = 0 and x/4 - (y-2)/2 = 0

To solve the system, it looks fairly easy to eliminate x:

Multiply the hyperbola equation by 16:
x^2 - 4(y-2)^2 = 16
Subtract this from the circle equation:
5y^2 - 16y + 16 = 84
5y^2 - 16y - 68 = 0

OK, I've nothing clever to offer here. Use the formula to find y, each of the two values, when substituted in the circle equation, should give two values for x, corresponding to four points of intersection (or two, or none, depending on whether you get real roots or not).
Actually, you'll see from a rough sketch of the graphs that all roots should be real, as there are four points of intersection.

Answer 2

The first is a circle with center (0,0) and radius 10
[Thats from the general equation of a circle which is x^2 + y^2 = r^2 where r = radius] ... Basically graph your circle with a radius of 10-units from your origin (0,0).

The second one is a hyperbola with a center at (0,2) and opens to the left and to the right (horizontal parallel to the x-axis). You can sketch a rectangle to help you find a sketch of the graph using the lengths "a" and "b" from the general equation: x^2/a^2 - y^2/b^2. For your problem these correspod to a^2 = 16 and b^2 = 4, therefore a=4 and b=2. Now you start at your center (0,2) and count 4 units to the right and to the left [by the way these points are also your vertices], as well as 2 units up and 2 units down. This will give you the "edges" of the rectangle from which you connect its vertices to find the asymptotes of your hyperbola. Now graph your hyperbola start from your vertices (4,2) and (-4,2) and sketch it so that it opens to the right and left and along your asymptotes.

To solve these equations you can look at the graph and wherever they intersect, then that is your solution.

Also, you can solve these equations algabraically as follows:

Take the first equation and solve for x^2 ... you get:

x^2 = 100 - y^2

Now substitute that into your second equation for just the x^2 value... you get:

(100-y^2) / 16 - (y-2)^2 / 4 = 1 [note: you dont square the first binomial since it already represents x^2]

Now FOIL both binomials which need squaring and solve to get the following trinomial in terms of "y"

-5y^2 + 16y + 68 = 0

Now use the quadratic formula to get your solutions for "y". Now plug that back in into your formulas to find your solutions to "x". Since you have "squares" don't forget that when you square-root something you have 2 solutions, one is + and the other is -

(e.g. sqroot of 9 = +3 or -3)

Both solutions should be taken into account, giving you 4 intersection points, i.e., 4 solutions to your problem.

Good luck!

Below is a link with examples of what your hyperbola should look like, except that it will have a different center and might open up wider or narrower in your case.

Answer 3

The graphing part is easy because all you have is a circle and an ellipse. The harder part is finding the intersections. I would solve the first equatio for x^2: x^2=100-y^2 and substitute this into the second and solve.

Answer 4

Something does not add up. I got this far...
x^2=100-y^2

take 2nd equation and x 16, we have
x^2 - 4(y-2)^2=16
x^2- 4(y^2 -4y +4)=16
x^2-4y^2+16y-16=16
(100-y^2)-4y^2+16y-32=0
68+16y-5y^2=0
I am stuck here--sorry