The equation of the tangent to a curve $y = f(x)$ at a point where $x = a$ and $y = f(x)$ at which the derivative $f'(x)$ exists and is finite, is,
$$\frac{y-f(a)}{x-a}=f'(a)$$
Find the equation to the tangent for the following curve $y^2 = 4ax$, normal, sub-tangent and sub-normal.
So what I have tried for tangent was the following:
$$y = \sqrt{4a^2}$$ because $x=a$, then plugging this into the equation above: $$\frac{y - 2a}{x-a} = 2 \\ y-2a = 2x-2a \\ y = 2x$$
Then for the normal, instead of $f'(a)$ I use instead $-\frac{1}{f'(a)}$, this gives me $$\frac{y-2a}{x-a}=-\frac{1}{2} \\ y-2a = -\frac{x-a}{2}$$
However, the solutions tell me I should have $x-y+a=0$ for tangent and $x+y=3a$ for normal, respectively.
For sub-tangent I have $\frac{f(a)}{f'(a)}$ and sub-normal I get $f(a)f'(a)$
Therefore,
sub-tangent $ = \frac{2a}{2} = a$, sub-normal $ = 2a \times 2 = 4a$, however, these are shown to be $(2a, 2a)$ respectively according to solutions.
Much help is appreciated in correcting the maths!