text: precalc help 5? x^2 - 2y^2 = 16  what is the center, vertices, foci and asymptotes of the hyperbola divide both sides by 16 to get x^2/a^2 - y^2/b^2 = 1, which is the standard form. since x^2 is positive, that tells you that this hyperbola opens horizontally. because you just have a  plain x^2 and y^2, the center is (0,0). the square root of the number under the x^2 term gives you the distance of the vertices from the center, so the vertices are at (a,0) and (-a,0). the asymptotes are lines with the slope formed by +/- b/a. to find the foci, calculate c = sqrt(a^2 + b^2) and go that distance from the center horizontally, so the foci are at (c,0) and (-c,0). i leave it to you to do the math, but that's the process for this problem.