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Sketch the curve (y^2 - 2)^2 + (x^2 - 2)^2 = 2

I have no idea about how to approach this question:

Sketch th curve (y^2 - 2)^2 + (x^2 - 2)^2 = 2 what would it look like?
Thanks :smile:

Reply 1

Farhan.Hanif93

Original post by Mundu348

I have no idea about how to approach this question:

Sketch th curve (y^2 - 2)^2 + (x^2 - 2)^2 = 2 what would it look like?
Thanks :smile:

What does the graph of

(y-2)^2+(x-2)^2=2

look like? Now, if you replace y with y^2, what effect does this have on the graph? Then do the same for x^2, what effect does this have? Think about particular intervals which will be effected by this.

Reply 2

Tobedotty

I just checked this curve using wolfram alpha, and im going to be honest, it is very very weird. But like farhan says, just think about the equation for a circle, then think about what happens if you use x^2s and y^2s instead.

Reply 3

Farhan.Hanif93

Original post by Mundu348

I have no idea about how to approach this question:

Sketch th curve (y^2 - 2)^2 + (x^2 - 2)^2 = 2 what would it look like?
Thanks :smile:

I think I may have been a bit vague as I don't think that you will get anywhere with that hint unless you got the idea yourself.
Note that

(y-2)^2+(x-2)^2=2 \implies y = 2\pm\sqrt{2-(x-2)^2}

.
Find the values of x for which y is less than 1. Then think about what would happen if you squared a number which is less than 1. What would happen to the values of y which are greater than 1 if you squared them? Don't forget to reflect the function in the x-axis.

Reply 4

Gary Gong

Actually,the graph is weird,but i think we can do that .First the graph is symmetric,so we only need to draw one part of that.
first factorising,y^4-4y^2+4+x^4-4x^2+4=2, y^4-4y^2+3+x^4-4x^2+3=0 ,(y^2-1)(y^2-3)+(x^2-1)(x^2-3)=0,it is obvious that when y=+-1 y=+-root 3 x=+-1 x=+-root 3,it is true .so we plot these point ,and draw a curve which is through these points.the graph is completed.