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

1. anonymous

help..

2. anonymous

@ganeshie8

3. Loser66

Take derivative of the curve, then plug it into the tangent line equation.

4. Loser66

4x + 10yy'=0 hence $$y'= \dfrac{-4x}{10y}$$, at (8,-1), $$y'= 16/5$$

5. anonymous

4x+10y(dy/dx)=0 4(8)+10(-1)dy/dx=0?

6. anonymous

ok then wats next?

7. 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.

8. Loser66

where $$(x_0, y_0) =(8,-1)$$

9. ganeshie8

we need to be careful, (8, -1) is not on the ellipse

10. Loser66

Wow!! ok, let me check

11. Loser66

oh, yea, so, the tangent line to the curve which passes through (8,-1) right? @ganeshie8

12. anonymous

ok so y+1=16/5x-128/5 y=(16/5)x+128/5

13. Loser66

Sorry AliLnn. I misread the question. hehehe... I guided you wrong until the derivative. Let @ganeshie8 take over. @ganeshie8 Please.

14. anonymous

ok

15. ganeshie8

|dw:1443270407665:dw|

16. anonymous

ok then |dw:1443267223012:dw|like that..?

17. anonymous

hmm i have to go for a while but u can post.. and ill respond later ok

18. 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

19. mathmate

or better still, http://prntscr.com/8knmq8

20. 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

21. mathmate

@irishboy123 An excellent geometric approach, thank you!

22. anonymous

ok thanks to all ..i think i understand it now