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

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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...