How to derive the equation of tangent to an arbitrarily point on a ellipse? 
Show that the equation of a tangent in a point $P\left(x_0, y_0\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, could be written as: $$\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$$

I've tried implicit differentiation $\to \frac{2x\frac{d}{dx}}{a^2}+\frac{2y\frac{d}{dy}}{b^2} = 0$, but not sure where to go from here. Substituting $P$ doesn't seem to help me much, and solving for $y$ from original equation seems to cause me more trouble than help. Please don't give me the solution, rather just give a slight hint or two:)
Thanks in advance!
Edit:
Solving for $\frac{dy}{dx}$ and putting it into point slope gives me:
$\frac{d}{dx} \left[\frac{x^2}{a^2} + \frac{y^2}{b^2}\right] = \frac{d}{dx}\left[1\right] \to \frac{2x}{a^2} + \frac{y}{b^2}\frac{dy}{dx}= 0\to \frac{dy}{dx}=-\frac{xb^2}{ya^2} \to \frac{dy}{dx}(P)=-\frac{x_0b^2}{y_0a^2}$ 
Then we get:
$y-y_0=\frac{dy}{dx}\left(x-x_0\right)\to y = -\frac{x_0b^2}{y_0a^2}(x-x_0)+y_0\to y = \frac{x_0^2b^2}{y_0a^2}-\frac{xx_0b^2}{y_0a^2} + y_0\to /:b^2,*y_0\to \frac{yy_0}{b^2}+\frac{xx_0}{a^2}=\frac{y_0^2}{b^2}$
Which looks close, but not exactly the expression i wanted. Where is the error?
 A: This probably isn't the kind of answer you're looking for, but it's one I like because it doesn't use calculus.
The ellipse is the image of the unit circle under the linear mapping $T(x,y) = (ax,by)$. The tangent line to the ellipse is the image under $T$ of the tangent line to the circle.
We have $n = T^{-1}(x_0,y_0) = (x_0/a,y_0/b)$, the corresponding point on the unit circle. The equation of the tangent line to the circle at $n$ is $1 = n \cdot (x,y) = xx_0/a + yy_0/b$.
To get the equation of the tangent line to the ellipse at $T(n) = (x_0,y_0)$, we substitute $T^{-1}(x,y) = (x/a,y/b)$ into this equation to obtain
$$1 = xx_0/a^2 + yy_0/b^2.$$ 
A: Use the fact that the gradient of a differentiable function at a point is orthogonal to the level set of that function passing through that point.
Here,
$$f(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2},$$
and
$$\vec\nabla f(x_0,y_0)=\frac{2x_0}{a^2}+\frac{2y_0}{b^2}.$$
Hence, for $(x_0,y_0)\in\mathbb{R}^2$ such that $f(x_0,y_0)=1$, an equation of the tangent line to the ellipse $f(x,y)=1$ at $(x_0,y_0)$ is
$$\vec\nabla f(x_0,y_0)\cdot(x-x_0,y-y_0)=0,$$
i.e.,
$$\frac{2x_0(x-x_0)}{a^2}+\frac{2y_0(y-y_0)}{b^2}=0,$$
i.e.,
$$\frac{2x_0x}{a^2}+\frac{2y_0y}{b^2}=\frac{2x_0^2}{a^2}+\frac{2y_0^2}{b^2}.$$
The result follows after dividing by $2$ and using the fact that $f(x_0,y_0)=1$.
A: Let the parametric equation of the tangent be
$$x=x_0+t\cos(\theta),y=y_0+t\sin(\theta),$$ where $\theta$ is unknown.
Plug in the equation of the ellipse to get
$$\frac{(x_0+t\cos(t))^2}{a^2}+\frac{(y_0+t\sin(t))^2}{b^2}=1\\
=\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}+2\left(\frac{x_0\cos(\theta)}{a^2}+\frac{y_0\sin(\theta)}{b^2}\right)t+\left(\frac{\cos^2(\theta)}{a^2}+\frac{\sin^2(\theta)}{b^2}\right)t^2.$$
After simplification, this equation is of the form $\alpha t^2+\beta t=0$. As the line is tangent, the root $t=0$ must be double. This occurs if $\beta=0$,
$$\frac{x_0\cos(\theta)}{a^2}+\frac{y_0\sin(\theta)}{b^2}=0.$$
Then for any $t$
$$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=\frac{(x_0+t\cos(\theta))x_0}{a^2}+\frac{(y_0+t\sin(\theta))y_0}{b^2}=\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}=1.$$
A: The answer is given in Perspectives on Projective Geometry by Jürgen Richter-Gebert (credit to @Jan-MagnusØkland for this insight):

A conic consists of all points p that satisfy an equation $p^TAp = 0.$ The set
of all tangents to this conic can be described as $\{Ap | p^TAp = 0\}.$

The proof is that since $\forall p,q \in \mathbb R^3,p \cdot q := p^T q$ and $p \cdot q = 0$ implies orthogonality, the equations $q = Ap$ define tangent lines (in line coordinates) to the conic at points $p$.
Thus, in general, the tangent lines to any conic (not just ellipses) can be described by
$$
\left(\matrix{x & y & 1}\right) A \left(\matrix{ x_0 \\ y_0 \\ 1 }\right) 
$$
A: To show that the given line is tangent to the ellipse, one can show that the system of equations
\begin{align}
\frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1 \tag{I} \\
\frac{xx_0}{a^2} + \frac{yy_0}{b^2} &= \tag{II} 1
\end{align}
with $x$, $y$ variables has exactly one solution—namely $(x_0, y_0)$. We further assume that $(x_0, y_0)$ lies on the ellipse, i.e.,
$$ \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1. \tag{III} $$
Taking $(\mathrm I) - 2(\mathrm{II}) + (\mathrm{III})$ produces
$$ \frac{x^2 - 2xx_0 + x_0^2}{a^2} + \frac{y^2 - 2yy_0 + y_0^2}{b^2} = 0 $$
or
$$ \left(\frac{x-x_0}{a}\right)^2 + \left(\frac{y-y_0}{b}\right)^2 = 0. $$
The only way for the sum of two squares to be zero is that $x=x_0$ and $y=y_0$, hence this is indeed the only solution as desired.
A: We first differentiate the ellipse equation w.r.t. $x$  to find the slope of the tangent $m$,
$$\dfrac{2x_0}{a^2}+\dfrac{2y_0m}{b^2}=0$$
$$m=-\dfrac{x_0}{y_0}\cdot\dfrac{b^2}{a^2}$$
Thus, the tangent is given by
$$\color{darkgreen}{y-y_0}=\color{blue}{-}\dfrac{\color{blue}{x_0}}{\color{darkgreen}{y_0}}\cdot\dfrac{\color{darkgreen}{b^2}}{\color{blue}{a^2}}\color{blue}{(x-x_0)}$$
$$\color{darkgreen}{\dfrac{yy_0}{b^2}-\dfrac{y_0^2}{b^2}}=\color{blue}{\dfrac{x_0^2}{a^2}-\dfrac{xx_0}{a^2}}$$
$$\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}=\dfrac{x_0^2}{a^2}+\dfrac{y_0^2}{b^2}=1$$
as desired.
Hope this helps. :)
