Use implicit differentiation to find an equation of the tangent line to the curve at the given point 
Use implicit differentiation to find an equation of the tangent line to the curve $$x^2+xy+y^2=1$$ at  $(1,1)$.

I am not sure how I should work this out because the given point is not on the curve.
 A: [1] $x^2 + xy + y^2 = 1$
differentiating implicitly by $\frac{d}{dx}$ 
$2x + x\frac{dy}{dx} + y + 2y \frac{dy}{dx} = 0$
Which simplifies to
[2] $\frac{dy}{dx} = \frac{-2x - y}{x + 2y}$
Now consider a tangent linear equation which passes through a point on the curve and through $(1,1)$
It has the form 
[3] $y = mx + c$
To find the points that these equations [1] and [3] intersect we must solve them simultaneously.
Substituting $y$ into [1]
$x^2 + x(mx+c) + (mx+c)^2 = 1$ 
This simplifies to 
$(m^2+m+1)x^2 + (2mc+c)x + c^2 = 1$ 
using the quadratic formula 
[4] $x = \frac{-2mc-c \pm \sqrt{(2mc+c)^2 - 4(m^2+m+1)c^2}}{2(m^2+m+1)}$
Note that there are 2 solutions because there are two points on the curve where this will work.
Also, since we know that equation [3] passes through the point $(1,1)$
we know
$1= m1+c$
So [5] $m = 1 - c$
Se expression [4] becomes
[6] $x = \frac{-2(1 - c)c-c \pm \sqrt{(2(1 - c)c+c)^2 - 4((1 - c)^2+(1 - c)+1)c^2}}{2((1 - c)^2+(1 - c)+1)}$
Lets call these $x_1$ and $x_2$ They have a corresponding $y_1$ and $y_2$ 
We also know that the gradient of [3] is $m$. This will be equal to the derivative given by [2] at $(x_1, y_1)$ and $(x_2,y_2)$
ie:
[7] $m = \frac{-2x_1 - y_1}{x_1 + 2y_1} = 1 - c$
and
[8] $m = \frac{-2x_2 - y_2}{x_2 + 2y_2} = 1 - c$
Substitute [6] into [7] to get an expression for $y_1$ in terms of c
Then you can substitute $x_1$ and $y_1$ into [1] to find what $c$ is.
From here you can calculate $m$ which will give you the tangent line you are looking for.
A similar process can be used for $x_2$ and $y_2$ there are two tangent lines which will solve this problem. 
I'll leave the rest for you
A: Given $\ x^2 + xy + y^2 = 1$, find tangent line thru (1,1)
$\ Solve \ for \ y $
$ y = \frac{-x \pm \sqrt {4-3x^2}}{2}= \frac{-x}{2}\pm\frac{\sqrt{4-3x^2}}{2}=f(x)$
$\frac{dy}{dx}=\frac{-1}{2}\mp\frac{1.5x}{\sqrt{4-3x^2}}=f'(x)$
Let point (a, f(a)) is tangent point; for the slope, we have the equation
$\ f'(a)=\frac{f(a)-1}{a-1}$
Solve this for a , we obtain
$ a \approx -0.483 \ ; f'(a) \approx -0.10118$
The tangent line y = f '(a)(x - 1) + 1
$y = -0.10118(x - 1) + 1$
By inverse symmetry, also $y = -0.483(x - 1) + 1$
desmos pic
