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anonymous
 one year ago
Two lines through the point (1,3) are tangent to the curve x^2+4y^24x8y+3=0. find the equation.
anonymous
 one year ago
Two lines through the point (1,3) are tangent to the curve x^2+4y^24x8y+3=0. find the equation.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have been workin on this last night but can't get the right slope

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the correct slope is suppose to be 11/4 and 11/44 but can't get it to be like that. I used implicit differentiation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Hero @markaskingalexandria1 @m3m3man1999 @MahoneyBear @MathHater82 @m&msdoodle @marigirl @e.s.b @Empty @e.v.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0pls don't tag me ;;

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here is what i have done so far but the slopes i got are 11/2 etc...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@m&msdoodle @Hero @theopenstudyowl @pebbles_xx3 @PlasmaFuzer @plohrb @plzzhelpme @pooja195 @proud_yemeniah_ @pacely.chitlen.98

Hero
 one year ago
Best ResponseYou've already chosen the best response.0Calculus is not necessarily needed for this.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the only point given is (1,3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the equation given is a conic and it is an ellipse

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how would you solve it sir?

Hero
 one year ago
Best ResponseYou've already chosen the best response.0I take that back. At first I thought it was a circle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0any hints to get the slope? which is the dy/dx?

marigirl
 one year ago
Best ResponseYou've already chosen the best response.0would you happen to have the original question? could u post it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Two lines through the point (1,3) are tangent to the curve x^2+4y^24x8y+3=0. find the equation.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1@RAY_OMEGA you should get \[\Large \frac{dy}{dx} = \frac{2x+4}{8y8} = \frac{x+2}{4y4}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah my dy/dx is x2/4(y1)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Let point P be the point (1,3) Let Q be some point (x,y) on the elliptical curve Find the slope of line PQ \[\Large m = \frac{y_2y_1}{x_2x_1}\] \[\Large m = \frac{y3}{x(1)}\] \[\Large m = \frac{y3}{x+1}\]  set that slope equal to dy/dx \[\Large m = \frac{y3}{x+1}\] \[\Large \frac{dy}{dx} = \frac{y3}{x+1}\] \[\Large \frac{x+2}{4(y1)} = \frac{y3}{x+1}\] at this point I'm stuck. I have a feeling you have to solve `x^2+4y^24x8y+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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry for no reply I had to attend class

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So here is what I did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0After 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hey guys i already solved the problem here just gonna leave this open if you want to know the correct answer
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