SOLUTION: what are the foci of the graph x^2/36-y^2/16=1

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Question 460534: what are the foci of the graph x^2/36-y^2/16=1
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
The directrix is x=5 and the focus is (11, -7)
x%5E2%2F36-y%5E2%2F16=1
The vertex, by definition, is equidistant from the directrix and the focus. Since x+=+5 and x+=+11 the midpoint would be at x+=+8. So the vertex is at (8,+-7)
If you graph those points you will see the parabola must be on it's side, open to the right.
%28y+%2B+7%29%5E2+=+12%28x+-+8%29 would be the standard form equation.

Since 36 is under the x term, the major axis runs parallel to the x+-axis, and is 12 units long.
36+=+a%5E2
a+=+6
the major axis is 2a long.
Similarly, the minor axis runs parallel to the y-axis and is 8+units long.
If you want to find the location of the foci:
sqrt%2836-16%29
=sqrt+%2820%29
=+/-%282+sqrt%28+5%29%29
so the foci are located at about (-4.47,+0) and (4.47, 0)