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Solving Linear-Nonlinear Systems

  1. Apr 2, 2015 #1
    1. The problem statement, all variables and given/known data

    In a chapter in a book I'm studying from, they introduce quadratic inequalities (all conics, not just parabolas) but don't go over how to deduce which portion of the graph should be shaded. Considering there's two variables involved in these inequalities, I've been having trouble understanding without someone guidance.

    I've included an image of one of their example problems of the two that were provided in the book.

    ImageUploadedByPhysics Forums1428006616.322451.jpg

    2. Relevant equations

    Circle has one standard form only, as for the rest the equation that's on a horizontal axis centered at origin and the second equation is on the vertical axis can be achieved by interchanging x and y. To manipulate the center, replace x^2 with (x-h)^2 and y^2 with (y-k)^2 (except for the parabola).
    Standard Form of a Circle: x^2+y^2=r^2 (radius)
    Standard Form of an Ellipse: x^2/a + y^2/b = 1
    Standard Form of a Parabola: y = a(x-h)^2+k
    Standard Form of a Hyperbola: x^2/a - y^2/b = 1

    3. The attempt at a solution

    After simplifying the second equation into it's standard form for example, I ended up with:

    x^2/16 - y^2/4 > 1

    I understand what the ellipse looks like and how to graph it and everything — but I don't understand what the "> 1" implies. I can understand that it means shade inside the conic section from the graph they've provided — but what does this imply for circles? Other hyperbolas?

    I'm sorry, I'm completely lost on this chapter.
    Sincerely, Andrew
     
  2. jcsd
  3. Apr 2, 2015 #2

    haruspex

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    I'm not sure I understand what it is that you don't understand... what you are asking seems too simple.
    You write that you understand what "the ellipse" looks like.... did you mean the hyperbola?
    Having sketched the conic, all you need to do is find which side of the curve matches the inequality, right? Find a point on an axis that is clearly in one region, substitute the x and y values in the equation, and see if the inequality is true.
     
  4. Apr 2, 2015 #3

    Ray Vickson

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    If f(x,y) contains linear, quadratic and constant terms, you can usually determine the region f(x,y) >= 0 by checking if the origin (x,y)=(0,0) satisfies the inequality or not. The only time this fails is when the boundary f = 0 passes through the origin. In that case, test another point instead, such as (x,y ) = (1,0) or something similar.
     
  5. Apr 2, 2015 #4

    LCKurtz

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    @Etheryte: I may be being redundant with this but... I suppose it goes without saying that the graph of ##f(x,y)=0## is the locus of all points where the function is zero. That means everywhere else it is either true that ##f(x,y)>0## or ##f(x,y)<0##. So the graph divides the plane into those types of regions. Quadratics are all continuous so the function can't go from positive to negative anywhere but on the graph. That's why you only need to check a point in each region as others have suggested.
     
  6. Apr 2, 2015 #5

    SteamKing

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    If you have a circle defined by the equation x2 + y2 = r2, then this circle is centered at the Origin and has radius r.

    When you take a point (x0, y0) and plug its value into the equation for this circle:

    if x02 + y02 = r2, the point (x0, y0) lies on the circle

    if x02 + y02 < r2, the point (x0, y0) lies inside the circle

    if x02 + y02 > r2, the point (x0, y0) lies outside the circle
     
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