# precalc

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What is the transverse axis of the following equation?
25x^2+250x-36y^2-576y=2579

• precalc -

25(x^2 + 10x) - 36(y^2 + 16y) = 2579
Complete the squares:

25(x^2 + 10x + 25) - 25*25 - 36(y^2 + 16y + 64) + 36*64 = 2579

25(x+5)^2 - 36(y+8)^2 = 900

How convenient . . .

(x+5)^2/36 - (y+8)^2/25 = 1

Looks like an hyperbola with center at (-5,-8) and horizontal transverse axis

y = -8.

• precalc -

Thank you steve

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