Prove that the equation of the tangent at P is $ \frac {xx_1}{a^2} - \frac {yy_1}{b^2} = 1 $ (Hyperbolas) 

Question: Point P ($x_1 , y_1$) is on the hyperbola $\frac {x^2}{a^2}$ - $\frac {y^2}{b^2}$ = 1
Prove that the equation of the tangent at P is 
$$ \frac {xx_1}{a^2} - \frac {yy_1}{b^2} = 1 $$



What I have attempted:
Equation of a tangent is:
$$ (y-y_0) = m(x-x_0) $$
In this case we have $x_0 = x_1$ and $y_0 = y_1$ Now we need to find $m$ so I am going to implicitly differentiate the equation of the hyperbola
$$ \frac {x^2}{a^2} - \frac {y^2}{b^2} = 1 $$
$$ \frac {2x}{a^2} - \frac {2y}{b^2} (\frac {dy}{dx})= 1 $$
$$ \frac {dy}{dx}= \frac {xb^2}{ya^2}$$
at $x = x_1$ and $y = y_1$
$$ \frac {dy}{dx}= \frac {x_1b^2}{y_1a^2}$$
With $ (y-y_0) = m(x-x_0) $
$$ (y-y_1) = \frac {x_1b^2}{y_1a^2}(x-x_1) $$
$$ (y-y_1) = \frac {xx_1b^2}{y_1a^2} -  \frac {x_1^2b^2}{y_1a^2} $$
$$ (y-y_1) = \frac {xx_1b^2 - x_1^2b^2 }{y_1a^2}  $$
$$ yy_1a^2 - y_1^2a^2 = xx_1b^2 - x_1^2b^2 $$
$$ \frac{yy_1}{b^2} -\frac{y_1^2}{b^2}  = \frac{xx_1}{a^2}  - \frac{x_1^2}{a^2}$$
$$ \frac{x_1^2}{a^2} -\frac{y_1^2}{b^2} = \frac{xx_1}{a^2} - \frac{yy_1}{b^2} $$
Now I am stuck I am trying to get rid of 
$ \frac{x_1^2}{b^2} -\frac{y_1^2}{b^2} $ and try to make that equal to $1$ and prove the statement but the equation of the hyperbola is given $\frac {x^2}{a^2}$ - $\frac {y^2}{b^2}$ = 1...

Since $(x_1,y_1)$ lie on hyperbola, they satisfy the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$
$$ \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2} = 1$$
$$\implies\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$$
