## anonymous one year ago Two lines through the point (-1,3) are tangent to the curve x^2+4y^2-4x-8y+3=0. find the equation.

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1. anonymous

I have been workin on this last night but can't get the right slope

2. anonymous

the correct slope is suppose to be -11/4 and -11/44 but can't get it to be like that. I used implicit differentiation.

3. anonymous

@Hero

4. anonymous

@Hero @markaskingalexandria1 @m3m3man1999 @MahoneyBear @MathHater82 @m&msdoodle @marigirl @e.s.b @Empty @e.v.

5. anonymous

pls don't tag me ;-;

6. anonymous

7. anonymous

8. anonymous

here is what i have done so far but the slopes i got are -11/2 etc...

9. anonymous

@m&msdoodle @Hero @theopenstudyowl @pebbles_xx3 @PlasmaFuzer @plohrb @plzzhelpme @pooja195 @proud_yemeniah_ @pacely.chitlen.98

10. Hero

Calculus is not necessarily needed for this.

11. anonymous

but the only point given is (-1,3)

12. anonymous

the equation given is a conic and it is an ellipse

13. anonymous

how would you solve it sir?

14. Hero

I take that back. At first I thought it was a circle.

15. anonymous

any hints to get the slope? which is the dy/dx?

16. marigirl

@jim_thompson5910

17. marigirl

@misty1212

18. marigirl

would you happen to have the original question? could u post it?

19. anonymous

Two lines through the point (-1,3) are tangent to the curve x^2+4y^2-4x-8y+3=0. find the equation.

20. jim_thompson5910

@RAY_OMEGA you should get $\Large \frac{dy}{dx} = \frac{-2x+4}{8y-8} = \frac{-x+2}{4y-4}$

21. anonymous

Yeah my dy/dx is x-2/4(y-1)

22. jim_thompson5910

Let point P be the point (-1,3) Let Q be some point (x,y) on the elliptical curve Find the slope of line PQ $\Large m = \frac{y_2-y_1}{x_2-x_1}$ $\Large m = \frac{y-3}{x-(-1)}$ $\Large m = \frac{y-3}{x+1}$ ------------------------------------------------------- set that slope equal to dy/dx $\Large m = \frac{y-3}{x+1}$ $\Large \frac{dy}{dx} = \frac{y-3}{x+1}$ $\Large \frac{-x+2}{4(y-1)} = \frac{y-3}{x+1}$ at this point I'm stuck. I have a feeling you have to solve x^2+4y^2-4x-8y+3=0 for y, then replace each y with that expression in terms of x, but that looks like it will get very messy real fast.

23. anonymous

24. anonymous

So here is what I did

25. anonymous

After I got the dy/dx of the given curve I used at point slope formula plugging in (-1,3) as the points x and y and dy/dx asy slope

26. anonymous

Hey guys i already solved the problem here just gonna leave this open if you want to know the correct answer