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anonymous
 one year ago
Need Help! Find the points on the ellipse x^2 + 2y^2 =1 where the tangent line has slope 1.
anonymous
 one year ago
Need Help! Find the points on the ellipse x^2 + 2y^2 =1 where the tangent line has slope 1.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Did you try implicit differentiation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea, that's what I was thinking to do. But after I did that, what would I do?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439244740487:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now what do I do bubbas

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 2x }{ 4y }=1\] x=2y \[x^2+2y^2=1\] \[4y^2+2y^2=1\] 6y^2=1 \[y=\pm \frac{ 1 }{ \sqrt{6} }\] find x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I just plug the y values into the ellipse equation to find the x value?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no x=2y remember x and y are of opposite sign.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can also put in the eq. of ellipse.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So what am I doing wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im right, I matched the answer in the book

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you plug in eq. of ellipse you may write wrong sign but in this you automatically get the correct sign

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How would I get the wrong sign, if I get negative and positive

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2+2*\frac{ 1 }{ 6 }=1,x^2=1\frac{ 1 }{ 3 }=\frac{ 2 }{ 3 },x=\pm \sqrt{\frac{ 2 }{ 3 }}\] now while writing points one may match wrong sign

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x^2 + 2y^2 = 1 2x + 2*2y * y' = 0 x + 2y * y' = 0 y ' = x/(2y) solve y' = 1 x/(2y) = 1 y = x/2 x^2 + 2(x/2)^2 = 1 x^2 + 2* x^2 / 4 = 1 x^2 + x^2/2= 1 3/2 x^2 = 1 x^2 = (2/3) x = +  √2/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's a good idea to confirm this with a graphing calculator that can plot implicitly defined curves in R^2

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0another way \(x^2 + 2y^2 =1\) is an **ellipse** around origin as in: \(\large (\frac{x}{a})^2 + (\frac{y}{b})^2 = 1\) \( x = a \ cos \theta, \ y = b \ sin \theta \) \( \implies a = 1 \ \ .. \ b = \frac{1}{\sqrt{2}}\) tangent line [in rectangular] \(\large = \frac{dy}{dx} = \frac{d(a \ cos \ \theta)}{d(b \ sin \ \theta)} = \frac{a \ sin \theta \ d \theta}{b \ cos \theta \ d \theta} \implies \frac{a \ sin \theta}{b \ cos \theta } = 1\) \(\large \implies tan \theta = \frac{b}{a} = \frac{1}{\sqrt{2}}\) dw:1439250471290:dw \(\large \implies x = cos \theta = \pm \sqrt{\frac{2}{ 3}}, \ \ ..\ y = \frac{1}{\sqrt{2}} sin \theta = \frac{1}{\sqrt{2}}\times \mp \frac{1}{\sqrt{3}} = \mp \frac{1}{\sqrt{6}}\)
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