Finding Extreme x-Values on a Parabola

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SUMMARY

The discussion focuses on finding the extreme x-values of the parabola defined by the equation 9x² + 24xy + 16y² + 20x - 15y = 0, which has its vertex at the origin. Participants clarify that due to the presence of the "xy" term, the parabola is angled and may be bounded in both x and y directions. The correct approach involves implicit differentiation, treating x as a function of y, leading to the derivative dx/dy = (15 - 32y) / (18x + 44). Setting the denominator to zero identifies critical points for extreme x-values.

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Homework Statement



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

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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
 
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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.
 
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.
 
Thanks guys; I implicitly differentiated and found that \frac {dx}{dy} = \frac {15 - 32y}{18x + 44}. 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?
 
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.
 
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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