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!
\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}