Question

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

Answer 1

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

Answer 2

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.

Answer 3

@Hero

Answer 4

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

Answer 5

pls don't tag me ;-;

Answer 6

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

Answer 7

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

Answer 8

Calculus is not necessarily needed for this.

Answer 9

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

Answer 10

the equation given is a conic and it is an ellipse

Answer 11

how would you solve it sir?

Answer 12

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

Answer 13

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

Answer 14

@jim_thompson5910

Answer 15

@misty1212

Answer 16

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

Answer 17

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

Answer 18

@RAY_OMEGA

you should get \[\Large \frac{dy}{dx} = \frac{-2x+4}{8y-8} = \frac{-x+2}{4y-4}\]

Answer 19

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

Answer 20

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

\[\Large m = \frac{y_2-y_1}{x_2-x_1}\]

\[\Large m = \frac{y-3}{x-(-1)}\]

\[\Large m = \frac{y-3}{x+1}\]

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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.

Answer 21

Sorry for no reply I had to attend class

Answer 22

So here is what I did

Answer 23

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

Answer 24

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