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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
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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Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Take derivative of the curve, then plug it into the tangent line equation.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.14x + 10yy'=0 hence \(y'= \dfrac{4x}{10y}\), at (8,1), \(y'= 16/5\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.04x+10y(dy/dx)=0 4(8)+10(1)dy/dx=0?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1that 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
 one year ago
Best ResponseYou've already chosen the best response.1where \((x_0, y_0) =(8,1)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1we need to be careful, (8, 1) is not on the ellipse

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Wow!! ok, let me check

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1oh, yea, so, the tangent line to the curve which passes through (8,1) right? @ganeshie8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so y+1=16/5x128/5 y=(16/5)x+128/5

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Sorry AliLnn. I misread the question. hehehe... I guided you wrong until the derivative. Let @ganeshie8 take over. @ganeshie8 Please.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443270407665:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok then dw:1443267223012:dwlike that..?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm i have to go for a while but u can post.. and ill respond later ok

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3Let the tangent line has a slope of m If it passes through (8,1), then y(1)=m(x8) => y=m(x8)1...........(1) Given the ellipse: 2x^2+5y^2=70 we substitute y from (1) into the ellipse equation: 2x^2+5((m(x8)1)^2)=70 => (5m^2+2)x^2(80m^2+10m)x+(320m^2+80m65)=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)^24*(5m^2+2)*(320m^2+80m65)=0 simplifying and factoring, 40*(m+1)(29m13)=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
 one year ago
Best ResponseYou've already chosen the best response.3or better still, http://prntscr.com/8knmq8

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\(2x^2 + 5y^2 = 70\) \(\vec n = <2x, 5y>\) from point (8,1), \(\vec t = <8x, 1y>\) \(\vec n \bullet \vec t = 0 \implies 16x  2x^25y  5y^2 = 0\) \(\implies 16x  5y = 70\) [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+16x5y%3D70

mathmate
 one year ago
Best ResponseYou've already chosen the best response.3@irishboy123 An excellent geometric approach, thank you!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thanks to all ..i think i understand it now
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