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

1. anonymous

Did you try implicit differentiation?

2. anonymous

Yea, that's what I was thinking to do. But after I did that, what would I do?

3. anonymous

|dw:1439244740487:dw|

4. anonymous

Now what do I do bubbas

5. IrishBoy123

solve it

6. IrishBoy123

y' = 1

7. 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

8. anonymous

oh okay let me see

9. anonymous

So I just plug the y values into the ellipse equation to find the x value?

10. anonymous

no x=-2y remember x and y are of opposite sign.

11. anonymous

you can also put in the eq. of ellipse.

12. anonymous

So what am I doing wrong

13. anonymous

Im right, I matched the answer in the book

14. anonymous

Thanks either way

15. anonymous

if you plug in eq. of ellipse you may write wrong sign but in this you automatically get the correct sign

16. anonymous

How would I get the wrong sign, if I get negative and positive

17. 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

18. 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

19. anonymous

It's a good idea to confirm this with a graphing calculator that can plot implicitly defined curves in R^2

20. 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| $$\large \implies x = cos \theta = \pm \sqrt{\frac{2}{ 3}}, \ \ ..\ y = \frac{1}{\sqrt{2}} sin \theta = \frac{1}{\sqrt{2}}\times \mp \frac{1}{\sqrt{3}} = \mp \frac{1}{\sqrt{6}}$$