# Implicit function

1. Oct 14, 2007

1. The problem statement, all variables and given/known data
a. Find dy/dx given that x$$^{2}$$=y$$^{2}$$-4x=7y=15
b. under what conditions on x and/or y is the tangent line to this curve horizontal? vertical?

2. The attempt at a solution

I did solve the first question by simply using implicit fuction.

2x+2y*y'-4+ty*y'=0
y'=$$\frac{4-2x}{9y}$$

above is the answer that i ended up with

but im kinda stuck with the second question
if the tangent line is horizonat i guess the slope has to be zero
so i just set $$\frac{4-2x}{9y}$$=0 and
i ended up with x=2
but i don't know what to do with "vertical" part and
i'm not sure if i got the "horizonal" part either.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 14, 2007

### bob1182006

for a that's right, just when you wrote out the question you forgot some +'s and put = instead.

yes horizontal tangent is when the derivative =0. so x=2 is current for that.
vertical tangent occurs when the the derivative = +-infinity.

and if you have a fraction when does it = infinity?

3. Oct 14, 2007

### NonAbelian

If the question is supposed to be x²+y²-4x+7y = 15 then I believe that your answer for y' is incorrect.

Note that your original equation is that of a circle which has been shifted. To write it in standard form, (x-a)²+(y-b)²=c² you can complete the square (I'm getting to a nice geometrical result, just hang on :D)

(x²-4x+4) -4 + (y²+7y + 49/4) - 49/4 = 15
(x-2)² + (y+7/2)² = 15 +4 + 49/4
(x-2)² + (y+7/2)² = [sqrt(125)/2]²

So your original equation is a circle of radius [sqrt(125)/2], centered at (2,-7/2). From this you should gather that there are 2 points that have a horizontal tangent, and two points that have a vertical one.