In the system below what are the coordinates of the solution that lies in quadrant II?

x^2 + 4y^2 = 80
y = 1/32x^2

Question

In the system below what are the coordinates of the solution that lies in quadrant II?

x^2 + 4y^2 = 80
y = 1/32x^2

anonymous · Answer

First, sketch them out to get a rough idea. |dw:1394410839857:dw|

I drew a box around the point of intersection in the second quadrant. We don't care about the function in the other 3 quadrants, so let's find a way to describe the part of the circle we care about in terms of x.

$$4y^2=80-x^2$$

$$y=\pm \sqrt{20-\frac{ 1 }{ 4 }x^2}$$

In the part of the graph we care about, is y positive or negative? It's positive, so we use the positive square root. Set it equal to the other equation, 1/32 x^2, to find out the x value for where they intersect. 

$$y=\ \sqrt{20-\frac{ 1 }{ 4 }x^2}$$

...continued