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anonymous

  • one year ago

Find the equation of the tangent through the external point (8,-1) to the ellipse 2x^2+5y^2=70

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  1. anonymous
    • one year ago
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    help..

  2. anonymous
    • one year ago
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    @ganeshie8

  3. Loser66
    • one year ago
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    Take derivative of the curve, then plug it into the tangent line equation.

  4. Loser66
    • one year ago
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    4x + 10yy'=0 hence \(y'= \dfrac{-4x}{10y}\), at (8,-1), \(y'= 16/5\)

  5. anonymous
    • one year ago
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    4x+10y(dy/dx)=0 4(8)+10(-1)dy/dx=0?

  6. anonymous
    • one year ago
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    ok then wats next?

  7. Loser66
    • one year ago
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    that is m, right? that is the slope of the tangent line at (8,-1), y - y_0 = m (x -x_0) , that is it.

  8. Loser66
    • one year ago
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    where \((x_0, y_0) =(8,-1)\)

  9. ganeshie8
    • one year ago
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    we need to be careful, (8, -1) is not on the ellipse

  10. Loser66
    • one year ago
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    Wow!! ok, let me check

  11. Loser66
    • one year ago
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    oh, yea, so, the tangent line to the curve which passes through (8,-1) right? @ganeshie8

  12. anonymous
    • one year ago
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    ok so y+1=16/5x-128/5 y=(16/5)x+128/5

  13. Loser66
    • one year ago
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    Sorry AliLnn. I misread the question. hehehe... I guided you wrong until the derivative. Let @ganeshie8 take over. @ganeshie8 Please.

  14. anonymous
    • one year ago
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    ok

  15. ganeshie8
    • one year ago
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    |dw:1443270407665:dw|

  16. anonymous
    • one year ago
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    ok then |dw:1443267223012:dw|like that..?

  17. anonymous
    • one year ago
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    hmm i have to go for a while but u can post.. and ill respond later ok

  18. mathmate
    • one year ago
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    Let the tangent line has a slope of m If it passes through (8,-1), then y-(-1)=m(x-8) => y=m(x-8)-1...........(1) Given the ellipse: 2x^2+5y^2=70 we substitute y from (1) into the ellipse equation: 2x^2+5((m(x-8)-1)^2)=70 => (5m^2+2)x^2-(80m^2+10m)x+(320m^2+80m-65)=0.......(2) We solve the quadratic (2) with the proviso that there is only one point of contact, i.e. the discriminant=0, This means (80m^2+10m)^2-4*(5m^2+2)*(320m^2+80m-65)=0 simplifying and factoring, -40*(m+1)(29m-13)=0 => m=-1 or m=13/29 Substitute m into the above equation (1) will give the two tangent lines, as required. See http://prntscr.com/8knlfx for confirmation of results

  19. mathmate
    • one year ago
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    or better still, http://prntscr.com/8knmq8

  20. IrishBoy123
    • one year ago
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    \(2x^2 + 5y^2 = 70\) \(\vec n = <2x, 5y>\) from point (8,-1), \(\vec t = <8-x, -1-y>\) \(\vec n \bullet \vec t = 0 \implies 16x - 2x^2-5y - 5y^2 = 0\) \(\implies 16x - 5y = 70\) [as \((2x^2 + 5y^2 = 70\)] solve for \(2x^2 + 5y^2 = 70\) & \(\ 16x - 5y = 70\) http://www.wolframalpha.com/input/?i=solve+2x%5E2+%2B+5y%5E2+%3D+70%2C+16x-5y%3D70

  21. mathmate
    • one year ago
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    @irishboy123 An excellent geometric approach, thank you!

  22. anonymous
    • one year ago
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    ok thanks to all ..i think i understand it now

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