anonymous
  • anonymous
Find the equation of the tangent through the external point (8,-1) to the ellipse 2x^2+5y^2=70
Mathematics
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anonymous
  • anonymous
Find the equation of the tangent through the external point (8,-1) to the ellipse 2x^2+5y^2=70
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
help..
anonymous
  • anonymous
Loser66
  • Loser66
Take derivative of the curve, then plug it into the tangent line equation.

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Loser66
  • Loser66
4x + 10yy'=0 hence \(y'= \dfrac{-4x}{10y}\), at (8,-1), \(y'= 16/5\)
anonymous
  • anonymous
4x+10y(dy/dx)=0 4(8)+10(-1)dy/dx=0?
anonymous
  • anonymous
ok then wats next?
Loser66
  • Loser66
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.
Loser66
  • Loser66
where \((x_0, y_0) =(8,-1)\)
ganeshie8
  • ganeshie8
we need to be careful, (8, -1) is not on the ellipse
Loser66
  • Loser66
Wow!! ok, let me check
Loser66
  • Loser66
oh, yea, so, the tangent line to the curve which passes through (8,-1) right? @ganeshie8
anonymous
  • anonymous
ok so y+1=16/5x-128/5 y=(16/5)x+128/5
Loser66
  • Loser66
Sorry AliLnn. I misread the question. hehehe... I guided you wrong until the derivative. Let @ganeshie8 take over. @ganeshie8 Please.
anonymous
  • anonymous
ok
ganeshie8
  • ganeshie8
|dw:1443270407665:dw|
anonymous
  • anonymous
ok then |dw:1443267223012:dw|like that..?
anonymous
  • anonymous
hmm i have to go for a while but u can post.. and ill respond later ok
mathmate
  • mathmate
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
mathmate
  • mathmate
or better still, http://prntscr.com/8knmq8
IrishBoy123
  • IrishBoy123
\(2x^2 + 5y^2 = 70\) \(\vec n = <2x, 5y>\) 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
mathmate
  • mathmate
@irishboy123 An excellent geometric approach, thank you!
anonymous
  • anonymous
ok thanks to all ..i think i understand it now

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