i need help in algebra 2.?

Question

i know how to do this kinda, but its still really hard. write the parabola equation in standard form.x=5y^2-20y+15what is the standard form for this: x^2-5y^2-8x-9=0 is it a circle, ellipse, or hyperbolax^2+y2+12x-26y+61=0 a parabola cirlce, ellipse, or hyperbola. 4.21x^2-84x+12y^2-24y=6 a parabola, circle, ellipse, or hyperbola

Mr. Ellis Hintz · Accepted Answer

1) x=5y^2-20y+15 >> This is already in the standard form with Horizontal axis of symmetry:x = ay^2 + by + c And in the vertex form :(y - k)^2 = 4p(x - h)2) x^2-5y^2-8x-9=0 >> hyperbola:3)x^2+y2+12x-26y+61=0 >> circle4)21x^2-84x+12y^2-24y=6 >> ellipse(x - h)2 / a2 + (y - k)2 / b2 = 1

Mr. Ellis Hintz · Answer

standard form of a line is just get all the variables on the left, and the numbers on the right x = 5y^2 - 20y +15x - 5y^2 +20y = 15 is standard form2.x^2 - 5y^2 - 8x = 9dont get too caught up in forms. You have to learn it to answer questions, but it distracts you from what is really going on.This is a hyperbola. since both y and x are squared you know it is not a line or parabola. If it was a circle or elipse they would be added. Since they are subtracted, it is a hyperbolalets simplify it into propper elipse form by completing the squarex^2 - 8x - 9 -5y^2 = 0x^2 - 8x -5y^2 = 9x^2 - 8x +(-8/2)^2 -5y^2 = 9 + (-8/2)^2 x^2 - 8x +16 - 5y^2 = 25(x-4)^2 - 5y^2 = 25(x-4)^2/25 - 5y^2/25 = 1this is standard form of a hyperbola3.complete the square for both x and yx^2 + 12x + y^2-26y = -61(x^2 + 12x + 36) + (y^2 - 26y + 13^2) = -61 + 36 + 169(x+6)^2 + (y-13)^2 = 144(x+6)^2/144 + (y-13)^2/144 = 1 is a circle 12 diameter, center at (-6,13)4.4.21(x^2 - 84/4.21x) + 12(y^2 -2y) = 64.21(x^2 - 84/4.21x + [(84/4.21)/2]^2) + 12(y^2 -2y +1) = 6 + 4.21[(84/4.21)/2]^2 +12(1)4.21(x - 84/8.42)^2 + 12(y-1)^2 = 6 + 4.21(84/8.42)^2 +124.21(x - 84/8.42)^2 / (18 + 4.21(84/8.42)^2) + 12(y-1)^2 / (18 + 4.21(84/8.42)^2)dont let all the decimals scare you, just keep track of everything and be careful

Mr. Ellis Hintz · Answer

What you need is something called completing the square. This video should help you out: http://khanexercises.appspot.com/video?v=LSJuu4Qm2... .Answers: 1) Parabola, x = 5 (y - 2)^2 - 352) Hyperbola, 1 = ((x-4)^2) / 25 - (y^2) / 53) Circle, 144 = (x+6)^2 + (y-13)^24) Ellipse, 1 = (7(x-2)^2) / 34 + (6(y-1)^2) / 51

Mr. Ellis Hintz · Answer

1)y1 = 2 + (1/5)sqrt(5x + 25), y2 = 2 - (1/5)sqrt(5x + 25)2)y1 = (1/5)sqrt(5x^2 - 40x - 45), y2 = - (1/5)sqrt(5x^2 - 40x - 45)3)y1 = 13 + sqrt(108 - x^2 - 12x), y2 = 13 - sqrt(108 - x^2 - 12x)4) y1 = 1 + (1/2)*sqrt(6 - 7*x^2 + 28*x), y2 = 1 - (1/2)*sqrt(6 - 7*x^2 + 28*x)

Mr. Ellis Hintz · Answer

3) is a circle i think lol. maybe

Prof. Kayli VonRueden

i need help in algebra 2.?

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