Equation of a ParabolaDate: 03/04/97 at 16:05:40 From: croman family Subject: Parabolas as quadratic functions Dr. Math, I'm a 10th grade student and I'm having some trouble in my honors algebra II class with parabolas as quadratic functions. Could you tell me how to do these problems, please: 1. Determine the equation of a parabola with a vertex at the origin and a focus of (10,0). 2. Determine the equation of a parabola with a vertex at the origin and a directrice of y = -2 If you know any way to make this stuff easier, please tell me. If it is at all possible, please show your work. The entire class will be grateful. Sincerely, April Croman Date: 03/06/97 at 16:54:12 From: Doctor Keith Subject: Re: Parabolas as quadratic functions Hi! First let's show how we got that a parabola is a quadratic equation. Remember that a quadratic equation is of the form: y = f(x) = Ax^2 + Bx + C A parabola is the set of points in the plane that are equally distant from a point (focus) and a line (the directrix). We want to write this as an equation, so let's define our notation and assume for the moment that our directrix is a horizontal line (i.e., the parabola opens up or down): focus is the point (a,b) point of parabola (x,y) directrix is the line y = d distance euclidean distance d(pt1,pt2) = sqrt((x1-x2)^2 + (y1-y2)^2) With these definitions, we know that the distance from the focus to any point on the parabola is: d((a,b),(x,y)) = sqrt((a-x)^2 + (b-y)^2) The distance from the directrix to a point on the parabola is: d((x,d),(x,y))= sqrt((x-x)^2 + (d-y)^2) = | d-y | Now we set them equal to each other and square both sides to clean things up: (a-x)^2 + (b-y)^2 = (d-y)^2 We need to collect terms so we will expand the ones that deal with y first: (a-x)^2 + y^2 - 2by + b^2 = y^2 - 2dy + d^2 Take the expanded terms all to the right side to cancel y^2 terms: (a-x)^2 = 2(b-d)y + d^2 - b^2 Now expand the left side: x^2 -2ax + a^2 = 2(b-d)y + d^2 - b^2 Now solve for y: x^2 -2ax + a^2 + b^2 - d^2 = 2(b-d)y (1/2(b-d))x^2 + (a/(d-b))x +((a^2)/2(b-d) + (b+d)/2) = y Let's make the following definitions: A = 1/2(b-d) B = a/(d-b) C = (a^2)/2(b-d) + (b+d)/2 Then we have our quadratic equation: y = Ax^2 + Bx + C I will leave the analogous derivation for parabolas opening left or right to you. Now we can solve your problems quickly by using this derivation and the fact that the vertex is a point on the parabola and that the directrix is perpendicular to the line traveling through the focus and the vertex: 1) We know (a,b) = (10,0) and vertex = (0,0). Thus the directrix is avertical line (perpendicular to x axis) and ten units away from the origin in the negative direction (for the equal distance part). So the directrix is x = -10. Then substitute these values in the equation you were to derive above for A, B, and C. 2) We know y = -2 is the directrix so the parabola opens up or down (perpendicular requirement) and since the vertix is the origin, it must open up and the focus is two units up on the y-axis (equal distance on opposite side) so (a,b) = (0,2). You can now substitute these values into the equation I derived above for A, B, and C and you are done. Hope this helps, If you would like more explanation, or something is confusing, write back and I will give you more. I left you some work so you could test yourself. -Doctor Keith, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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