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$x^2+y^2+cx^2y^2=1$

1- What happens to the curve when $c=-1$? Describe what appears on the screen. Can you prove it algebraically?

2- Find $y'$ by implicit differentiation. For the case $c=-1$, is your expression for $y'$ consistent with what you discovered in part (b)?

I've solved the first question and found that the curve is two parallel lines ($y=1$, $y=-1$); and proved that algebraically. My proof is as follows, please correct me if I'm wrong:

When $c=-1, x^2+y^2-x^2y^2=1$. Then, $y^2-x^2y^2=1-x^2$. Then, $y^2(1-x^2)=1-x^2$. Dividing both sides by $(1-x^2)$, $y^2=1, y=1 or -1$ when $x\ne1$

The problem is in the 2nd question. The derivative of that curve when $c=-1$ is $\left[ y'= \dfrac{y^2x-x}{y(1-x^2)}\right]$, but according to the solution of the first question the slope must be ($0$), how can we make the two solutions consistent with each other???

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  • $\begingroup$ The locus of points satisfying $x^2+y^2-x^2y^2=1$ is not a pair of parallel lines. It is $x=1\lor x=-1\lor y=1\lor y=-1$. $\endgroup$
    – symplectomorphic
    Aug 11, 2017 at 4:09
  • $\begingroup$ Can you explain further? or how can we solve the entire two questions? $\endgroup$
    – Abdu Magdy
    Aug 11, 2017 at 4:12
  • $\begingroup$ I can explain further, yes. But you should try again, knowing that your answer to (1) was incorrect. $\endgroup$
    – symplectomorphic
    Aug 11, 2017 at 4:13
  • $\begingroup$ How can I message you; so that I can show you my proof $\endgroup$
    – Abdu Magdy
    Aug 11, 2017 at 4:16
  • $\begingroup$ If you have questions about your proof, put your proof in the question. $\endgroup$
    – symplectomorphic
    Aug 11, 2017 at 4:17

1 Answer 1

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When $c=-1$ then the curve reduces to $(y^2-1)(x^2-1)=0.$ That means either $y= 1$ or $y=-1$ or $x=1$ or $x=-1.$ So the graph consists of two vertical lines and two horizontal lines.

When $y=1$ or $y=-1$ then the expression you've got for $y'$ becomes $0,$ as you'd expect given that those are two horizontal lines. When $x=1$ or $x=-1,$ then the denominator in the expression you've got for $y'$ becomes $0$ and the derivative is undefined, as you'd expect for two vertical lines.

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  • $\begingroup$ I got it now. Thanks so much. I think the problem was because of Desmos Graphing Calculator. when you put the equation when, it shows only the horizontal lines. So I made it to be proven as two horizontal lines without noticing when $x = 1 or -1$... Any way, do you have a recommendation for a graphing calculator that shows the real graph, I mean without any mistake... @symplectomorphic, if you can help me in that as well... $\endgroup$
    – Abdu Magdy
    Aug 11, 2017 at 4:48

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