Show that the equation of the tangent to the parabola $y^2=4ax$ at the point (p,q) is $qy=2a(x+p)$ 

Question: Show that the equation of the tangent to the parabola $y^2=4ax$ at the point (p,q) is $qy=2a(x+p)$


These are my two approaches:
First approach:
If we have $(p,q)$ as $(x_1,y_1)$ 
$$y^2=4ax$$
$$ 2y \left( \frac{dy}{dx}\right) = 4a$$
$$ \frac{dy}{dx} = \frac{2a}{y} $$
At $y=q$
$$ \frac{dy}{dx} = \frac{2a}{q} $$
$$ m = \frac{2a}{q} $$
As equation of tangent is 
$$ (y-y_1) = m(x-x_1) $$
Plugging in values
$$ (y-q) = \frac{2a}{q}(x-p) $$
$$ y-q = \frac{2ax}{q} - \frac{2ap}{q} $$
$$ y-q = \frac{2ax-2ap}{q} $$
$$ qy - q^2 = 2ax-2ap $$
$$ qy -q^2 = 2a(x-p) $$
But this does not satisfy the proof

My second approach:
If $(p,q)$ are points on $y^2=4ax$ then 
plugging into $$ y^2 = 4ax $$
at $x=p$
$$ y^2 = 4ap $$
$$ y = \sqrt{4ap} $$
at $y=q$ 
$$ q^2 = 4ax $$
$$ x = \frac{q^2}{4a} $$
Finding gradient:
$$y^2=4ax$$
$$ 2y \left( \frac{dy}{dx}\right) = 4a$$
$$ \frac{dy}{dx} = \frac{2a}{y} $$
At $y = \sqrt{4ap} $
$$ \frac{dy}{dx}  = \frac{a}{\sqrt{ap}} $$
$$  (y-y_1) = m(x-x_1) $$
$$ y-\sqrt{4ap}=\frac{a}{\sqrt{ap}}(x-\frac{q^2}{4a})$$
But I don't think this gives me the proof as well...
 A: With your first approach is easier, I think. You got $\;y'=\frac{2a}y\;$ , so at point $\;(p,q)\;$ the slope is $\;\frac{2a}q\;$ , and thus the line with this slope and through this point is
$$y-q=\frac{2a}q(x-p)\iff qy-q^2=2a(x-p)$$
but $\;q^2=4ap\;$ since $\;(p,q)\;$  is on the parabola, and then:
$$qy-4ap=2ax-2ap\implies qy=2ax+2ap=2a(x+p)$$
A: My approach:
We have the equation of parabola : $y^2 = 4ax$ 
By taking the derivative of the above equation to get its gradient $m$, we have
$2y \left( \frac{dy}{dx}\right) = 4a$
Rearranging the above equation, 
$\left(\frac{dy}{dx}\right)$ = $\frac{4a}{2y}$ = $m$
The equation of tangent is : $(y-y_1) = m(x-x_1)$
Putting in value of $m$ here, we have
$(y-y_1) = \frac{4a}{2y} (x-x_1)$
Also we know that $(y_1,x_1)=(p,q)$. So substituting the values we get
$(y-p) = \frac{4a}{2y} (x-q)$
Now by multiplying $2y$ on both sides we get
$2y^2 - 2qy = 4ax - 4ap$ 
$2y^2 - 4ax + 4ap = 2qy$
Now here we can substitute the value of $y^2$ with $4ax$ (From the equation of the parabola)
By doing so we have,
$8ax - 4ax + 4ap = 2qy$
$4ax + 4ap = 2qy$
$4a(x + p) = 2qy$
$\color{green}{qy=2a(x+p)}$ 
A: In the first approach,
$$qy-q^2=2a(x-p)$$
But $(p,q)$ is a point on the parabola $y^2=4ax$.
Thus,
$$\tag1q^2=4ap$$
Substituting,
$$qy-4ap=2ax-2ap$$
$$qy=2a(x+p)$$

In the second approach,
$$y-\sqrt{4ap}=\frac{a}{\sqrt{ap}}(x-\frac{q^2}{4a})$$
From $(1)$,
$$\tag2\frac{q^2}{4a}=p$$
$$\tag3\sqrt{ap}=\frac{q}{2}$$
$$\tag4\sqrt{4ap}=q$$
Using $(2),(3)$ and $(4)$,
$$y-q=\frac{2a}{q}(x-p)$$
$$qy-q^2=2a(x-p)$$
which is the same equation as in the first case and reduces to
$$qy=2a(x+p)$$
