Find tangent line going through a point outside the curve in multivariable calculus Determine the equations for all lines tangent to the ellipse
$$ f(x,y)=x^2+xy+y^2=1 $$
and passing through the point (0,2)
I'm a bit stuck on this question.
I was thinking if I have an unknown point on the curve called (a.b). Then I can plug it in the curve.
$$f(a.b)=a^2+ab+b^2=1$$
And if I also find the derivative of it, which is:
$$(2a+b,a+2b) $$
But from here I really don't know how to proceed. Maybe if I can somehow set it up as simultaneous equations and find the unknown point, from there I can easily find the tangent line.
 A: Note that a parameterization for your ellipse is
$$ \begin{align} x(t)&= -\cos(t)+\frac{1}{\sqrt{3}}\sin(t)\\\\y(t)&= \cos(t)+\frac{1}{\sqrt{3}}\sin(t) \text{ .} \end{align}$$
The tangent (velocity) vector at any point would then be
$$ \vec{v} = \begin{bmatrix} x'(t)\\y'(t) \end{bmatrix} = \begin{bmatrix} \sin(t) - \frac{1}{\sqrt{3}}\cos(t)\\-\sin(t) - \frac{1}{\sqrt{3}}\cos(t) \end{bmatrix} \text{ .}$$
For a particular point along the parameterization $\vec{p} = (x(\tau),y(\tau))$ on the ellipse, the tangent line would be given by
$$ \vec{x} = \vec{p} + \vec{v}(\tau)t $$
where $\vec{x} = (x_\ell(t),y_\ell(t))$ is a parameterization for the tangent line.
Let me know if you have any follow-up questions.
A: Here is one approach:
$x^2+xy+y^2=1$
Taking derivative wrt. $x$, $2x + y + x y' + 2y y' = 0 \implies y' = -\frac{2x+y}{x+2y}$
This is the slope of the tangent line at a point $T (x, y)$ on ellipse. As the tangent line goes through point $P (0, 2)$, we can equate slope of $PT$ to $y'$.
$-\frac{2x+y}{x+2y} = \frac{y-2}{x} \implies x^2+xy+y^2 = x + 2y$.
As point $T (x, y)$ is on ellipse, $x^2 + xy + y^2 = 1$
So, $x = 1 - 2y$ and plugging it into equation of ellipse,
$(1-2y)^2 + y(1-2y) + y^2 = 1 \implies y(y-1) = 0$
That gives us two points on the ellipse $(1, 0)$ and $(-1, 1)$. Tangent lines at these points pass through point $P(0, 2)$.
You can now find the equation of tangent lines.
A: A general line through $(0,2)$ would be parametrised as $$\mathbf{r}(t)=(0,2)+t(\alpha,\beta).$$
Substitute that into the equation for the ellipse and then you will have an expression which you can treat as a quadratic in t. The line will be tangential to the ellipse if and only if there is just one solution for t i.e. discriminant $b^2-4ac=0$. This should narrow down the possible choices for the direction $(\alpha,\beta)$ to only two options. Hope this helps
A: Extremely easy using the trick of homogenization :
$$x^2 +y^2 + xy -1=0 \tag{1}$$
Homogenize.
$$ x^2 +y^2 + xy -z^2=0$$
Now, add in some mindless derivatives:
$$ X(2x +y) + Y(2y +x) +(-2z)=0$$
What does above represent? It represents of the chord of contact from the tangents drawn to an external point $P=(X,Y)$
Evaluate at z=1,
$$ X(2x+y) + Y(2y+x) - 2= 0 \tag{2}$$
Now, plug in $(0,2)=(X,Y)$
$$ 2(2y+x) -2 = 0$$
$$ 2y+x=1 \tag{3}$$
Now, simply find the intersection of the above equation with $(2)$, this will lead you to getting some $(x,y)$ pair which is a point on the curve. Each of this point corresponds to a tangent line you can draw from curve to external point, to convert the point to tangent line, plug the $(x,y)$ pair into $(2)$.
I find these pairs to be $(x,y) = \{ (-1,1), (1,0)\}$
Reference for method
