SOLUTION: The graph of the conic equation {{{16x^2+49y^2+192x=208}}} has vertices and foci equal to ??

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Question 1106060: The graph of the conic equation 16x%5E2%2B49y%5E2%2B192x=208 has vertices and foci equal to ??
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The equation has both x^2 and y^2 terms, with the same sign and different coefficients, so the equation is of an ellipse.

The standard form of the equation of an ellipse is

%28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1 if the major axis is in the x direction; or
%28x-h%29%5E2%2Fb%5E2+%2B+%28y-k%29%5E2%2Fa%5E2+=+1 if the major axis is in the y direction

In both formulas, (h,k) is the center of the ellipse; and a and b are the semi-major and semi-minor axes, respectively.

Two different forms of the equation are necessary, because for the ellipse the a has to be the length of the semi-major axis.

Parameter c is the distance from the center of the ellipse to each focus; for an ellipse, c%5E2+=+a%5E2-b%5E2. Note that this formula for finding the value of c is why it is necessary to have a > b.

To find the vertices and foci, you need to put the given equation in the standard form. To do that, you need to complete the square in both x and y, then divide by the appropriate constant to get the right side of the equation equal to 1.

16x%5E2%2B49y%5E2%2B192x=208
%2816x%5E2%2B192x%29%2B49y%5E2+=+208
16%28x%5E2%2B12x%29%2B49y%5E2+=+208
16%28x%5E2%2B12x%2B36%29%2B49y%5E2+=+208%2B16%2836%29+=+208%2B576+=+784
16%28x%2B6%29%5E2%2F784%2B49y%5E2%2F784+=+1
%28x%2B6%29%5E2%2F49+%2B+y%5E2%2F16+=+1
%28x%2B6%29%5E2%2F7%5E2+%2B+%28y-0%29%5E2%2F4%5E2+=+1

This is in standard form:
The center of the ellipse is (h,k) = (-6,0).
a is 7; the semi-major axis has length 7 in the x direction, so the vertices are at (-13,0) and (1,0).
b is 4; the semi-major axis has length 4 in the y direction. If you need the co-vertices, they are 4 units in the positive and negative y direction from the center -- at (-6,-4) and (-6,4).
The distance from the center to each focus is c, which is sqrt%28a%5E2-b%5E2%29+=+sqrt%2849-16%29+=+sqrt%2833%29. So the foci are at (-6-sqrt(33),0) and (-6+sqrt(33),0).