Implicit Line Finding?

**VonNemo19** · June 14th 2009, 10:50 AM

Find the equation of the line(s) tangent to the curve $x^3+2xy-y^2=-2$ at the point(s) where $x=1$ .

My answer is

$y=-\frac{3}{4}x+\frac{15}{4}$

and

$y=-\frac{5}{4}x+\frac{1}{4}$ .

Is this correct?

**HallsofIvy** · June 14th 2009, 11:38 AM

Originally Posted by VonNemo19

Find the equation of the line(s) tangent to the curve $x^3+2xy-y^2=-2$ at the point(s) where $x=1$ .

My answer is

$y=-\frac{3}{4}x+\frac{15}{4}$

and

$y=-\frac{5}{4}x+\frac{1}{4}$ .

Is this correct?

It would be better to post your work rather than just your answers.
But since you do: No, those are not correct at all. Post your work and we might be able to point out where you went wrong.

**VonNemo19** · June 14th 2009, 12:31 PM

Originally Posted by HallsofIvy

It would be better to post your work rather than just your answers.
But since you do: No, those are not correct at all. Post your work and we might be able to point out where you went wrong.

$x^3+2xy-y^2=-2$

Differentiating with respect to x...

$3x^2+2x*\frac{dy}{dx}+y*2-2y*\frac{dy}{dx}=0$

$(2x-2y)\frac{dy}{dx}=-3x^2-2y$

$\frac{dy}{dx}=\frac{-3x^2-2y}{2x-2y}$

Now, the points of tangecy are as follows

$(1)^3+2(1)y-y^2=-2$
$y^2-2y-3=0$
$(y-3)(y+1)=0\Rightarrow{y=3,-1}$

Therefore

$\frac{dy}{dx}\text{ at x=1 is }$

$\frac{-3(1)^2-2(3)}{2(1)-2(3)}=\frac{9}{4}$

and

$\frac{-3(1)^2-2(-1)}{2(1)-2(-1)}=\frac{-1}{4}$

So finding the equation...

Is this right so far? is there an easier way to do this?

**Jhevon** · June 14th 2009, 01:17 PM

Originally Posted by VonNemo19

$x^3+2xy-y^2=-2$

Differentiating with respect to x...

$3x^2+2x*\frac{dy}{dx}+y*2-2y*\frac{dy}{dx}=0$

$(2x-2y)\frac{dy}{dx}=-3x^2-2y$

$\frac{dy}{dx}=\frac{-3x^2-2y}{2x-2y}$

Now, the points of tangecy are as follows

$(1)^3+2(1)y-y^2=-2$
$y^2-2y-3=0$
$(y-3)(y+1)=0\Rightarrow{y=3,-1}$

Therefore

$\frac{dy}{dx}\text{ at x=1 is }$

$\frac{-3(1)^2-2(3)}{2(1)-2(3)}=\frac{9}{4}$

and

$\frac{-3(1)^2-2(-1)}{2(1)-2(-1)}=\frac{-1}{4}$

So finding the equation...

Is this right so far? is there an easier way to do this?

yes, you are correct so far. now what are the tangent lines for each point? (state the point as well)

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