Question 407754
classify the conic
x^2-4y^2+3x-26y-30=0

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x^2-4y^2+3x-26y-30=0
completing the squares
(x^2-3x+9/4)-4(y^2+(26/4)y+(169/16)=30+9/4-(169/4)
(x-3/2)^2-4(y+13/4)^2=-40/4=-10
multiply by (-1)
4(y+13/4)^2-(x-3/2)^2=10
divide by 10
((y+13/4)^2)/(10/4)-((x-3/2)^2)/10=1

ans: This is a hyperbola with centers at (3/2-13/4) and vertical transverse axis. Also, a^2=10/4 and b^2=10