Write down the equation of the tangent to $x^2-3y^2=4y$ at the point $(x_1,y_1)$. 
Write down the equation of the tangent to $x^2-3y^2=4y$ at the point $(x_1,y_1)$.

The textbook gives the answer as $xx_1-3yy_1=2(y+y_1)$ and I'm not sure how it got there.
Okay so I differentiated $x^2-3y^2=4y$ to get:
$\frac{dy}{dx}=\frac{x}{3y+2}$
Then to find the equation of the tangent I usee $y-y_1=m(x-x_1)$
$y-y_1=\frac{x_1}{3y_1+2}(x-x_1)$
So I'm unsure how  $y-y_1=\frac{x_1}{3y_1+2}(x-x_1)$ can be simplified to $xx_1-3yy_1=2(y+y_1)$ which is what the textbook says.
Help much appreciated.
 A: Your equation has a fraction, the one you want to get to does not. Your equation has parentheses, the textbook answer does not. 
Therefore, take your equation, multiply both sides by $3y_1+2$, expand the parentheses everywhere, and you should be a lot closer.
A: We differentiate
$$ x^2 - 3y^2 = 4y$$ 
To get, 
$$ 2x -6y \frac{dy}{dx} = 4 \frac{dy}{dx} $$ 
To arrive at,
$$ 2x = \frac{dy}{dx}(4 + 6y) $$ 
Yielding 
$$ \frac{x}{2 + 3y} = \frac{dy}{dx} $$ 
So at the target point as you rightly noted the slope will be 
$$ \frac{x_1}{2 + 3y_1} $$ 
The equation of the point then will be 
$$ \frac{x_1}{2 + 3y_1}(x - x_1) +y_1 = y $$ 
Which matches up with your solution. So now how to simplify this? We can multiply through by the denominator 
$$ x_1(x - x_1) + y_1(2 + 3y_1) = y(2 + 3y_1) $$ 
$$ xx_1 - x_1^2 + 2y_1 + 3y_1^2 = 2y + 3yy_1 $$ 
Recall that $$ x^2 - 3y^2 = 4y$$ from the equation and thus the term
$$ -x_1^2 + 3y_1^2 $$ can be rightly rewritten as $$-4y_1$$ 
$$ xx_1 - 4y_1 + 2y_1 = 2y + 3yy_1 $$ 
$$ xx_1 - 3yy_1 = 2y_1 + y $$
$$ xx_1 - 3yy_1 = 2(y + y_1) $$ 
The trick, is knowing to resubtitute the definition of the equation you got from the start. It's a bit subtle.
A: You got $y-y_1=\frac{x_1}{3y_1+2}(x-x_1)$ which is correct.
Now simplify $(y-y_1)(3y_1+2)=x_1(x-x_1)$ then $3yy_1-3y_1^2+2y-2y_1=xx_1-x_1^2$. By rearranging we get $xx_1-3yy_1-2y=x_1^2-3y_1^2-2y_1$. Adding $-2y_1$ both sides $\Rightarrow$ $xx_1-3yy_1-2(y+y_1)=x_1^2-3y_1^2-4y_1=0$ (Because $(x_1,y_1)$ is on the given curve $x^2-3y^2-4y=0$.) Hence the provided answer by your textbook. 
