Question 1106060
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The equation has both x^2 and y^2 terms, with the same sign and different coefficients, so the equation is of an ellipse.<br>
The standard form of the equation of an ellipse is<br>
{{{(x-h)^2/a^2 + (y-k)^2/b^2 = 1}}}  if the major axis is in the x direction; or
{{{(x-h)^2/b^2 + (y-k)^2/a^2 = 1}}}  if the major axis is in the y direction<br>
In both formulas, (h,k) is the center of the ellipse; and a and b are the semi-major and semi-minor axes, respectively.<br>
Two different forms of the equation are necessary, because for the ellipse the a has to be the length of the semi-major axis.<br>
Parameter c is the distance from the center of the ellipse to each focus; for an ellipse, {{{c^2 = a^2-b^2}}}.  Note that this formula for finding the value of c is why it is necessary to have a > b.<br>
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.<br>
{{{16x^2+49y^2+192x=208}}}
{{{(16x^2+192x)+49y^2 = 208}}}
{{{16(x^2+12x)+49y^2 = 208}}}
{{{16(x^2+12x+36)+49y^2 = 208+16(36) = 208+576 = 784}}}
{{{16(x+6)^2/784+49y^2/784 = 1}}}
{{{(x+6)^2/49 + y^2/16 = 1}}}
{{{(x+6)^2/7^2 + (y-0)^2/4^2 = 1}}}<br>
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(a^2-b^2) = sqrt(49-16) = sqrt(33)}}}.  So the foci are at (-6-sqrt(33),0) and (-6+sqrt(33),0).