# Extreme points on parabola

1. Aug 20, 2008

### avr10

1. The problem statement, all variables and given/known data

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

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. Aug 20, 2008

### HallsofIvy

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.

3. Aug 20, 2008

### cellotim

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.

4. Aug 20, 2008

### avr10

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?

5. Aug 20, 2008

### cellotim

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.

6. Aug 21, 2008

### cellotim

Set the derivative to 0 and solve for x or y. Then put back into the original equation to find the points.