anonymous
  • anonymous
Need Help! Find the points on the ellipse x^2 + 2y^2 =1 where the tangent line has slope 1.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Did you try implicit differentiation?
anonymous
  • anonymous
Yea, that's what I was thinking to do. But after I did that, what would I do?
anonymous
  • anonymous
|dw:1439244740487:dw|

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anonymous
  • anonymous
Now what do I do bubbas
IrishBoy123
  • IrishBoy123
solve it
IrishBoy123
  • IrishBoy123
y' = 1
anonymous
  • anonymous
\[-\frac{ 2x }{ 4y }=1\] x=-2y \[x^2+2y^2=1\] \[4y^2+2y^2=1\] 6y^2=1 \[y=\pm \frac{ 1 }{ \sqrt{6} }\] find x
anonymous
  • anonymous
oh okay let me see
anonymous
  • anonymous
So I just plug the y values into the ellipse equation to find the x value?
anonymous
  • anonymous
no x=-2y remember x and y are of opposite sign.
anonymous
  • anonymous
you can also put in the eq. of ellipse.
anonymous
  • anonymous
So what am I doing wrong
anonymous
  • anonymous
Im right, I matched the answer in the book
anonymous
  • anonymous
Thanks either way
anonymous
  • anonymous
if you plug in eq. of ellipse you may write wrong sign but in this you automatically get the correct sign
anonymous
  • anonymous
How would I get the wrong sign, if I get negative and positive
anonymous
  • anonymous
\[x^2+2*\frac{ 1 }{ 6 }=1,x^2=1-\frac{ 1 }{ 3 }=\frac{ 2 }{ 3 },x=\pm \sqrt{\frac{ 2 }{ 3 }}\] now while writing points one may match wrong sign
anonymous
  • anonymous
x^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
  • anonymous
It's a good idea to confirm this with a graphing calculator that can plot implicitly defined curves in R^2
IrishBoy123
  • IrishBoy123
another 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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