Finding the equation of tangent line I'm stuck with problem supposed to be trivial.
I need to find tangent line witch touches curve $y^2 = -4ax$ at the point $(x_0,y_0)$
Rewriting it as $$x = -\frac {y^2}{4a}$$
Taking derivative: $$f(y)'=-\frac {y}{2a}$$
Then slope is $$f(y_0)'=-\frac {y_0}{2a}$$
Then make a substitution in tangent line equation at the point $(x_0,y_0)$ to find the offset:
$$x = ky+b \iff b = x-ky \iff b = x_0 -\frac {y_0^2}{2a}$$
Then write our tangent line equation:
$$x = -\frac {y_0y}{2a} + x_0 -\frac {y_0^2}{2a} \iff 2ax = -y_0y + 2ax_0 -y_0^2$$
But this result doesn't match with result given in the book: $$4ax+2yy_0-y_0^2 = 0$$
Help me out, guys.
 A: You computed $\frac{dx}{dy}$ at $(x_0,y_0)$. This is the reciprocal of the slope. Easy fix, the slope at $(x_0,y_0)$ is $-\frac{2a}{y_0}$.
Alternately, differentiate implicitly. We get 
$$2y\frac{dy}{dx}=-4a.$$
So the slope of the tangent line at $(x_0,y_0)$ is $-\frac{4a}{2y_0}$. 
Remark: Your procedure should also have worked. However, there is a mistake in the next to last displayed line. It should be $2ax_0$, not $2x_0$. (When you cleared denominators by multiplying through by $2a$, the term $x_0$ did not get multiplied by $a$.)
Edit: The error has been corrected. To get the answer in the book, note that $2ax_0=-\frac{y_0^2}{2}$. 
A: There are a couple of mistakes in your working out (I see the second one has now been corrected in your question)
1. Where you write
$$b = x-ky \iff b = x_0 -\frac {y_0^2}{2a}$$
you should have instead
$$b = x-ky \iff b = x_0 +\frac {y_0^2}{2a} $$
2. In addition you forgot a factor of $a$ on the $x_0$ so
$$x = -\frac {y_0y}{2a} + x_0 +\frac {y_0^2}{2a} \iff 2ax = -y_0y + 2x_0 +y_0^2$$
becomes
$$x = -\frac {y_0y}{2a} + x_0 +\frac {y_0^2}{2a} \iff 2ax = -y_0y + 2ax_0 +y_0^2$$
Finally, you can just substitute for $2ax_0$ using $y_0^2=-4ax_0$, giving
$$ 2ax = -y_0y - \frac{y_0^2}{2} +y_0^2 $$
from which the result follows
$$ 4ax + 2y_0y - y_0^2 = 0 $$
