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

  2. anonymous
    • one year ago
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    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
    • one year ago
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    @Hero

  4. anonymous
    • one year ago
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    @Hero @markaskingalexandria1 @m3m3man1999 @MahoneyBear @MathHater82 @m&msdoodle @marigirl @e.s.b @Empty @e.v.

  5. anonymous
    • one year ago
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    pls don't tag me ;-;

  6. anonymous
    • one year ago
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  7. anonymous
    • one year ago
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  8. anonymous
    • one year ago
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    here is what i have done so far but the slopes i got are -11/2 etc...

  9. anonymous
    • one year ago
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    @m&msdoodle @Hero @theopenstudyowl @pebbles_xx3 @PlasmaFuzer @plohrb @plzzhelpme @pooja195 @proud_yemeniah_ @pacely.chitlen.98

  10. Hero
    • one year ago
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    Calculus is not necessarily needed for this.

  11. anonymous
    • one year ago
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    but the only point given is (-1,3)

  12. anonymous
    • one year ago
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    the equation given is a conic and it is an ellipse

  13. anonymous
    • one year ago
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    how would you solve it sir?

  14. Hero
    • one year ago
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    I take that back. At first I thought it was a circle.

  15. anonymous
    • one year ago
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    any hints to get the slope? which is the dy/dx?

  16. marigirl
    • one year ago
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    @jim_thompson5910

  17. marigirl
    • one year ago
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    @misty1212

  18. marigirl
    • one year ago
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    would you happen to have the original question? could u post it?

  19. anonymous
    • one year ago
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    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
    • one year ago
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    @RAY_OMEGA you should get \[\Large \frac{dy}{dx} = \frac{-2x+4}{8y-8} = \frac{-x+2}{4y-4}\]

  21. anonymous
    • one year ago
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    Yeah my dy/dx is x-2/4(y-1)

  22. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    Sorry for no reply I had to attend class

  24. anonymous
    • one year ago
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    So here is what I did

  25. anonymous
    • one year ago
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    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
    • one year ago
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    Hey guys i already solved the problem here just gonna leave this open if you want to know the correct answer

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