Implicit Diff.: 25x^2 + 16y^2 + 200x - 160y + 400 = 0

[LovesCalculus]

The problem is:

Find the points at which the graph of the equation below has horizontal tangent lines:

. . .25x^2 + 16y^2 + 200x - 160y + 400 = 0

What I did was solve for dy/dx and got:

. . .y`= -50x-200 / 32y-160

I do not know how to find the zeros of the function because there is a y in the equation. Normally, I would solve for the x's of the equation but I cannot do that. Any help would be greatly appreciated. Thanks SO much!

 

[stapel]

Hint: What is the value of the slope m for any horizontal line?

:wink:

Eliz.

 

[galactus]

I think they may be expecting you to notice this is the equation for an ellipse.

Group and complete the squares:

\(\displaystyle \L\\25x^{2}+200x+16y^{2}-160y=-400\)

\(\displaystyle \L\\25(x^{2}+8x+16)+16(y^{2}-10y+25)=400\)

Factor:

\(\displaystyle \L\\\frac{(x+4)^{2}}{16}+\frac{(y-5)^{2}}{25}=1\)