<s>[INST] Implicit differentiation of an equation of a hyperbola Prove that an equation of the tangent line to the graph of the hyperbola : 
$(x^2/a^2)  -  (y^2/b^2) = 1$
at the point ($x_0$, $y_0$) is 
$x x_0/a^2 - y y_0/b^2 = 1$  (1) 

I implicitly differentiated the equation and then found the gradient by substituting in the points to get the gradient ( $b^2x_0/a^2y_0$) and use the points, plug it into $y-y_1=m(x-x_1)$
But I don't know how to rearrange it to get to (1). Please help me!
 [/INST] \begin{align} 
y-y_0 & = \frac{b^2x_0}{a^2y_0}(x-x_0) 
\\ a^2y_0(y-y_0)& = (b^2x_0)(x-x_0) 
\\ a^2y_0y-a^2y_0^2& = b^2x_0x-b^2x_0^2 
\\ b^2x_0^2 -a^2y_0^2& = b^2x_0x-a^2y_0y
\end{align}
$(x_0,y_0)$ satisfaces that $\frac{x_0^2}{a^2}-\frac{y_0^2}{b^2}=1$ doing the operations, $b^2x_0^2 -a^2y_0^2=a^2b^2$, then:
\begin{align} 
\\ b^2x_0^2 -a^2y_0^2& = b^2x_0x-a^2y_0y
\\ a^2b^2& = b^2x_0x-a^2y_0y
\\ 1& = \frac{xx_0}{a^2}-\frac{yy_0}{b^2}
\end{align}
</s>