# SOLUTION: What are the correct vertices, foci, and asymptotes for the following hyperbola?: x^2-y^2+2x+6y-9=0 I completed the squares and ended up with this: (x+1)^2/(19) - (y-3)^2/(1

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: What are the correct vertices, foci, and asymptotes for the following hyperbola?: x^2-y^2+2x+6y-9=0 I completed the squares and ended up with this: (x+1)^2/(19) - (y-3)^2/(1      Log On

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 Question 629704: What are the correct vertices, foci, and asymptotes for the following hyperbola?: x^2-y^2+2x+6y-9=0 I completed the squares and ended up with this: (x+1)^2/(19) - (y-3)^2/(19)=1 But when I plug in (h-a,k)(h+a,k) the answer is not correct for the vertices: (-1-√19)(-1+√19) or (-1-√38,3)(-1+√38,3)for the foci. What am I doing wrong? Found 4 solutions by ewatrrr, Edwin McCravy, solver91311, lwsshak3:Answer by ewatrrr(10682)   (Show Source): You can put this solution on YOUR website!  Hi, x^2-y^2+2x+6y-9=0 (x+1)^2 -1 - (y-3)^2 +9 - 9 = 0 || In completing Square need to subtract 1 and add 9 C(-1,3) with V(-2,3) and V(1,3) Foci are ( -1 ± ,3) Standard Form of an Equation of an Hyperbola opening right and left is: with C(h,k) and vertices 'a' units right and left of center, 2a the length of the transverse axis Foci are units right and left of center along y = k & Asymptotes Lines passing thru C(h,k), with slopes m = ± b/a Answer by Edwin McCravy(8909)   (Show Source): You can put this solution on YOUR website!Apparently you didn't complete the square correctly: Half of 2 is 1. Squared is 1 Add +1 inside the first parentheses and to the right side. Half of -6 is -3. Squared is 9. Add +9 inside the second parentheses and since the second parentheses is preceded by a - sign we add -9 to the right side: That's equivalent to I think you can go from there. If not post again. Edwin Answer by solver91311(16897)   (Show Source): You can put this solution on YOUR website! Keeping the signs straight when completing the square on the general quadratic for a hyperbola can be a little tricky. Watch carefully: Move the constant term to the RHS: Rearrange: Since you are working a hyperbola, group the "negative variable" terms with parentheses: Note the sign change on the first degree term because of the distributive law. Now complete the square on both variables: Notice that because you are adding a +9 INSIDE of the parentheses on the left, you are actually adding a -9 to the LHS. So you need a -9 in the RHS to make up for it. . so center at just like you had it, but , , and therefore I think you can take it from here, can you not? John My calculator said it, I believe it, that settles it Answer by lwsshak3(6519)   (Show Source): You can put this solution on YOUR website!What are the correct vertices, foci, and asymptotes for the following hyperbola?: x^2-y^2+2x+6y-9=0 complete the squares x^2+2x-y^2+6y-9=0 (x^2+2x+1)-(y^2-6y+9)=9+1-9 (x+1)^2-(y-3)^2=1 (This is where you made a fatal error) This is an equation of a hyperbola with horizontal transverse axis. Its standard form:, (h,k)=(x,y) coordinates of center For given equation: center: (-1,3) a^2=1 a=1 vertices: (-1±a,-3)=(-1±1,-3)=(-2,-3) and (0,-3) .. b^2=1 b=1 .. c^2=a^2+b^2=1+1=2 c=√2≈1.4 Foci: (-1±c,-3)=(-1±1.4,-3)=(-2.4,-3) and (0.4,-3) .. Asymptotes are straight lines that go thru the center(-1,3) Standard form of equation for straight lines: y=mx+b, m=slope, b=y-intercept slopes of asymptotes for hyperbolas: ±b/a=±1/1=±1 Equations for asymptotes: y=x+b solve for b using coordinates of the center 3=-1+b b=4 equation: y=x+4 .. y=-x+b solve for b using coordinates of the center 3=1+b b=2 equation: y=-x+2