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Intersecting lines,circles,and parabolas

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    find the points in which the graphs intersect

    2. Relevant equations

    y=2x, x^2+y^2=1

    3. The attempt at a solution

    the center point of the circle is (0,0)
    and the radius = 1
    and i collect some points to draw the line equation
    (-1,-2) (0,0) (1,2) (2,4) etc ..

    but i don't know how to get the intersect points ..
    and also for the parabola equation ... are they have same way to know the intersect points ...
    is there a rule or what ??
    please explain to me in simple english words..

    thanx a lot ...
  2. jcsd
  3. Sep 24, 2011 #2


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    Homework Helper

    Just solve for x and y. Plug y = 2x into
    to get

    Solve for x (you should get two solutions), and plug both into
    y = 2x
    to get the corresponding solutions for y.

    The 2nd equation is an equation of a circle, not a parabola.
  4. Sep 24, 2011 #3


    User Avatar
    Science Advisor

    In simple English (nicht Deutsch, warum?), don't worry about the geometry, solve the equations!

    If the problem were to find the intersection of [itex]x^2+ y^2= 1[/itex] and [itex]y= x^2[/itex], the first is a circle and the second a parabola. That's nice to know (it tells us we can expect to find two points of intersection) but not necessary to the solution. Since [itex]y= x^2[/itex] we can replace [itex]x^2[/itex] by y in the first equation: [itex]y+ y^2= 1[/itex] or [itex]y^2+ y- 1= 0[/itex]. That's a quadratic equation and we can either complete the square or use the quadratic formula to solve for y. Once we have found y, x is a square root.

    (The quadratic equation has two roots, of course, and you might think that since each has two roots, there would be 4 (x, y) combinations. But one of the (real) roots to the quadratic is negative. That gives only imaginary roots and coordinates of points in a graph must be real. Only y> 0 gives the two points of intersection.)
  5. Sep 26, 2011 #4
    thanx alot ...
  6. Sep 26, 2011 #5
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