Answer Quadratic Implicit Function: Horizontal/Vertical Tangent Line

[ada0713]

Homework Statement

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 I am 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.

 

[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?

 

[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.