# Implicit Differentiation Graphs - help needed (1 Viewer)

#### blackops23

##### Member
Hi guys, here's the question I'm having trouble with:

5x^2 - y^2 + 4xy = 18 ---(1), I have to sketch it.

Here's what I did:
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10x - 2yy' + 4y + 4xy' = 0
5x - yy' + 2y + 2xy' = 0
y' = [(2y+5x)/(y-2x)]
As 2x-y--> 0, y'-->inf
therefore, y=2x, sub in ---(1)
5x^2 - 4x^2 + 8x^2 = 18
9x^2 = 18
x^2 = 2
x= +/- sqrt(2)
vertical tangents at [sqrt(2), 2*sqrt(2)] and [-sqrt(2), -2*sqrt(2)]

let numerator = 0
2y + 5x = 0
y= -5x/2, sub back in --- (1)
eventually...
-45x^2 = 72, therefore NO TURNING POINTS

Let x = 0, y^2= -18, no y-intercepts
let y=0, 5x^2 = 18
therefore x-intercepts: (sqrt(18/5), 0) and (-sqrt(18/5), 0 )
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So I've got two points with vertical tangents and 2 x-intercepts, what else can I do? Perhaps find any possible asymptotes? If so how do I do that?

Help greatly appreciated, thanks

#### Iruka

##### Member
Take eq(1) and complete the square:

5x^2 - y^2 + 4xy = 18

9x^2-(y^2-4xy+4x^2)=18
9x^2-(y-2x)^2=18

Using the difference of two squares, we have

(3x-y+2x)(3x+y-2x)=18
(5x-y)(x+y)=18

Now we can see that neither of the factors on the LHS can be zero since the RHS is not, so 5x-y=0 and x+y=0 gives you the equation of the two asymptotes.

There are probably other ways to do it, too.

#### blackops23

##### Member
thanks, but just wondering, are there any other methods to get the asymptotes?

#### Iruka

##### Member
You could express y as a function of x (or rather two functions of x, as you will have to take a square root somewhere) and then think real hard about what happens as x goes to infinity.