1. Find $a$ from the monotonicity of $f(x)=2x^2-\ln x$

We have $f'(x)=4x-\frac{1}{x}.$

For decreasing in $(0,a)$ and increasing in $(a,4)$, the critical point must be at $x=a$: $4x-\frac{1}{x}=0$ $4x^2=1$ $x=\frac{1}{2} \quad (x>0).$ Hence, $a=\frac{1}{2}.$

2. Equation of the parabola

Given $y^2=4ax,$ with $a=\frac12$, so $y^2=2x.$

A standard parametric point on $y^2=4ax$ is $P(at^2,2at).$ Since here $a=\frac12$, $P\left(\frac12 t^2,t\right).$

3. Equation of tangent at parameter $t$

For parabola $y^2=4ax$, tangent at parameter $t$ is $ty=x+at^2.$ With $a=\frac12$, $ty=x+\frac12 t^2.$

It passes through $(8a,8a-1)$. Since $a=\frac12$, $(8a,8a-1)=(4,3).$ Substitute $(x,y)=(4,3)$: $3t=4+\frac12 t^2$ $t^2-6t+8=0$ $(t-2)(t-4)=0.$ So, $t=2 \quad \text{or} \quad t=4.$

4. Use the condition that tangent does not pass through $\left(-\frac1a,0\right)$

Since $a=\frac12$, $\left(-\frac1a,0\right)=(-2,0).$

For tangent $ty=x+\frac12 t^2,$ if it passes through $(-2,0)$, then $0=-2+\frac12 t^2$ $t^2=4$ $t=\pm 2.$ Among the candidate values $t=2,4$, the value $t=2$ makes the tangent pass through $(-2,0)$, so it is rejected.

Hence, $t=4.$

Therefore, $P\left(\frac12\cdot 4^2,4\right)=(8,4).$

5. Equation of the normal at $P$

For $y^2=4ax$, differentiate: $2y\frac{dy}{dx}=4a$ $\frac{dy}{dx}=\frac{2a}{y}.$ At parameter $t$, since $y=2at$, slope of tangent is $m_t=\frac{1}{t}.$ So slope of normal is $m_n=-t.$

At $t=4$, normal slope is $m_n=-4.$ Passing through $(8,4)$, its equation is $y-4=-4(x-8)$ $y=-4x+36.$

Rewrite in intercept form: $4x+y=36$ $\frac{x}{9}+\frac{y}{36}=1.$ Thus, $\alpha=9,\quad \beta=36.$ So, $\alpha+\beta=45.$

6. Final answer

$\boxed{45}$

The derived answer matches the stored correct answer.