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.