1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook