A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

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

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Did you try implicit differentiation?

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1439244740487:dw|

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Now what do I do bubbas

  5. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    solve it

  6. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    y' = 1

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh okay let me see

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you can also put in the eq. of ellipse.

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So what am I doing wrong

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Im right, I matched the answer in the book

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks either way

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  20. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  21. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.