1. ## Implicit Line Finding?

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

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

and

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

Is this correct?

2. 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 $x=1$.

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

and

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

Is this correct?
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.

3. Originally Posted by HallsofIvy
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?

4. 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)