Question 812914
The standard forms for equations of ellipses are:
{{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}} for horizontally-oriented hyperbolas and
{{{(y-k)^2/a^2 - (x-h)^2/b^2 = 1}}} for vertically-oriented hyperbolas
Since the transverse axis is parallel to the x-axis in this problem, this hyperbola is horizontally-oriented. So we will be using the first form.<br>
In both forms, the coordinates of the center are represented by the 'h' and the 'k'. So with a center of (1, -2) out 'h' is 1 and our 'k' is -2.<br>
In both forms the 'a' represents the distance from the center to a vertex on the transverse axis. Since the center is halfway between the two vertices on the transverse axis, 'a' is 1/2 of the length of the transverse axis. This makes our 'a' 1/2 of 6 or, more simply, 3.<br>
With similar logic the 'b' is 1/2 the length of the conjugate axis. So our 'b' is 1/2 of 10 or just 5.<br>
With our 'h', 'k', 'a' and 'b' we are now ready to write the equation. Inserting the values we have found into {{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}} we get:
{{{(x-(1))^2/(3)^2 - (y-(-2))^2/(5)^2 = 1}}}
which simplifies to:
{{{(x-1)^2/9 - (y+2)^2/25 = 1}}}