The equation of such a hyperbola, with the transverse axis horizontal, is



where (h,k) is the center, a =  the length of the transverse axis,
and b =  the length of the conjugate axis.

We'll begin by drawing the horizontal line y = -5 and the vertical line x = 2.



The transverse axis and the conjugate axis intersect at the center of the
hyperbola which is (2,-5).  Now we'll leave just the transverse axis and the
conjugate axis, which are given as 6 units each, and we'll and erase the rest
of those green lines:

and draw the defining rectangle:

Now we can sketch in the asymptotes and the hyperbola:



We can write the equation of the hyperbola,



where (h,k) is the center (2,-5), a =  the length of the transverse
axis =  = 3
and b =  the length of the conjugate axis, also = 3





That is the equation in STANDARD, but the problem asks for it in the 
GENERAL form Ax² + Cy² + Dx + Ey + F = 0, so



Clear of fractions by multiplying through by 5

(x - 2)² - (y + 5)² = 9

x² - 4x + 4 - (y² + 10y + 25) = 9

x² - 4x + 4 - y² - 10y - 25 = 9

    x² - 4x - 21 - y² - 10y = 9

    x² - 4x - 30 - y² - 10y = 0

Rearrange the terms in the form Ax² + Cy² + Dx + Ey + F = 0

    x² - y² - 4x - 10y - 30 = 0


Edwin