<s>[INST] 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?
 [/INST] 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 $t$
$$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=\frac{(x_0+t\cos(\theta))x_0}{a^2}+\frac{(y_0+t\sin(\theta))y_0}{b^2}=\frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}=1.$$
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