Question 1173086
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{{{25x^2+9y^2+150x-36y+260 = 0}}} --> {{{25(x+3)^2+9(y-2)^2 = 1}}} --> {{{(x+3)^2/((1/5)^2)+(y-2)^2/((1/3)^2) = 1}}}<br>
In the denominators, 1/3 is greater than 1/5, so the semi-major axis a is 1/3 (in the y direction) and the semi-minor axis b is 1/5 (in the x direction).<br>
The vertices are 1/3 unit above and below the center; the co-vertices are 1/5 unit to the right and left of the center.<br>
The foci are c units above and below the center, where {{{c^2 = a^2-b^2 = (1/3)^2-(1/5)^2 = 1/9-1/25 = 25/225-9/225 = 16/225}}} so c = 4/15.<br>