At what point(s) on the curve x2 + y2 = 9 is the tangent line vertical?

Question

theslytherinhelper · Answer

I just clicked that answer and it won't go away, so please ignore it.

theslytherinhelper · Answer

I got as far as the derivative of the equation is (-x/y)?

loser66 · Answer

yup

theslytherinhelper · Answer

Great! I don't really know what to do after that, though. Do I have to plug in x values or set the equation to 0 or something?

loser66 · Answer

tangent line is vertical when y =0, replace it to original one to get $x=\pm 3$

theslytherinhelper · Answer

Okay, so, just to make sure, this is what I did (I'm not sure I did it right, but I did get +- 3.):
$$x^2+y^2 = 9$$
$$\sqrt(y^2) = 9- x^2$$
$$y=\sqrt(9-x^2)$$
$$0=\sqrt(9-9)$$
And then when you plug in 3 or -3, it gives you 0.

loser66 · Answer

It is not wrong but what you did doesn't relate to tangent line of a curve. You must take derivative and take limit of it to see how tangent = infinitive.

theslytherinhelper · Answer

So, would I just take the derivative $$\frac{ -x }{ y }$$, set the y to 0, and then plug in the 3?

loser66 · Answer

Need some verbal argument :)

loser66 · Answer

1) find dy/dx
2) limit dy/dx = 0 iff y =0
3) when y =0, x = +/- 3
4) conclusion.

theslytherinhelper · Answer

Thank you for your help!