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Extreme points on parabola

  1. Aug 20, 2008 #1
    1. The problem statement, all variables and given/known data

    The parabola [tex] 9x^{2} + 24xy + 16y^{2} + 20x - 15y = 0 [/tex] has vertex at the origin. Find the coordinates of the points on the parabola that have extreme x-values

    2. Relevant equations

    3. The attempt at a solution

    To start off, I don't quite understand the question. What does it mean for points on a parabola to have an extreme x-value? Don't parabolas extend til infinity in the x-direction? I thought about implicitly differentiating and going after the min/max values using the first derivative, but I don't see how that's going to answer the question. Someone have an idea? Thanks
  2. jcsd
  3. Aug 20, 2008 #2


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    Parabolas that have axis of symmetry parallel to the y-axis are unbounded in x but not y. Because of the "xy" term, this parabola is at an angle to the axes and may be bounded in both x and y. Since you want to find the "extreme x" values, you want to think of x as a function of y. Differentiate x with respect to y using implicit differentiation.
  4. Aug 20, 2008 #3
    It means find the points on the parabola corresponding to x maxima and minima. One of these is infinite. Go with implicit differentiation and find the points.
  5. Aug 20, 2008 #4
    Thanks guys; I implicitly differentiated and found that [tex]\frac {dx}{dy} = \frac {15 - 32y}{18x + 44} [/tex]. I used this to say that when the denominator is 0, namely when x is -44/18, x is an extreme. Does this sound right?
  6. Aug 20, 2008 #5
    Try doing the differentiation again. Remember you are taking the derivative w.r.t. y, assuming x is a function of y. Please don't forget to use the product rule.
  7. Aug 21, 2008 #6
    Set the derivative to 0 and solve for x or y. Then put back into the original equation to find the points.
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