Hey all. There's a question I seem to be stuck on involving implict differentiation. Here it is:(adsbygoogle = window.adsbygoogle || []).push({});

The curve called a bicorn has the equation [tex] (x^2+8y-16)^2=y^2(16-x^2) [/tex] Verify by implicit differentiation that its tangent lines at the points [tex] (0,4) [/tex] and [tex] (0,\frac{4}{3}) [/tex] are horizontal. Do by: a) expanding the equation

b) not expanding the equation

a) I expanded out the equation to get [tex] x^4+ x^2y^2 + 16x^2y-32x^2+48y^2-256y+256 [/tex] and took the derivative of all the terms and got the answer [tex] y'= \frac {-4x^3-2xy^2-32xy+64x}{2x^2y+16x^2+96y-256} [/tex]

b) Not expanding the equation and using the chain rule and product rule for each side respectively and then taking the derivative, I also got [tex] y'= \frac {-4x^3-2xy^2-32xy+64x}{2x^2y+16x^2+96y-256} [/tex]

My question is now,how would you solve the above derivative to get the points [tex] (0,4) [/tex] and [tex] (0,\frac {4}{3} ) [/tex] to verify it? Or would you just put in those two points in the derivative and assuming that I did it right, get the slope of 0? I'm pretty sure that the better way to do this is to solve for those two points since you could just put those points in the original equation [tex] (x^2+8y-16)^2=y^2(16-x^2) [/tex] to get zero but correct me if I'm wrong. So,any suggestions? Thank you for the help in advance.

P.S- Not that it really matters but here is a picture of a bicorn:

http://astronomy.swin.edu.au/~pbourke/curves/bicorn/1.gif [Broken]

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# Homework Help: Implicit differentiation and a bicorn curve

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