Let the function $\mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^2-{\log }_{\mathrm{e}}\mathrm{x},\mathrm{x}>0$ be decreasing in (0, a) and increasing in (a, 4). A tangent to the parabola $y^2=4ax$ at a  point P on it passes through the point (8a, 8a – 1) but does not pass through
the point $\left(-\frac{1}{a},0\right)$. If the  equation of the normal at P is $\frac{\mathrm{x}}{\mathrm{\alpha }}+\frac{\mathrm{y}}{\mathrm{\beta }}=1$, then $\mathrm{\alpha }+\mathrm{\beta }$ is equal to

$\Rightarrow \begin{array}{l}\frac{dy}{dx}=4x-\frac{1}{x}\\ =\frac{(2x+1)(2x-1)}{x}\\ \mathrm{f}\left(\mathrm{x}\right) \mathrm{decreases} \mathrm{in}(0,1/2) \mathrm{and} \\ \mathrm{f}\left(\mathrm{x}\right) \mathrm{increases} \mathrm{in} (1/2,\mathrm{\infty })\\ Equation of \mathrm{tangent} to y^2=4ax at \left(a{t}^2,2at\right)\\ \begin{array}{l}\mathrm{yt}=\mathrm{x}+{\mathrm{at}}^2\\ \text{ passes through }\left(8a,8a-1\right)\text{ =}(4,3)\\ \Rightarrow 3t=4+\frac{1}{2}{t}^2\\ t^2-6t+8=0\Rightarrow t=2,4\\ \begin{array}{l}\mathrm{t}=2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\mathrm{y}=\mathrm{x}+2\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ passes through }(-2,0)\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(-\frac{1}{a},0\right)=(-2,0)\\ \therefore \mathrm{t}=2\text{ not possible }\\ \begin{array}{r}\begin{array}{lll}t=4\Rightarrow Equation of \tan gent at P(8,4) is& 4y=x+8& \end{array}\\ \text{ Equation of normal at }(8,4) isy-4=-4(x-8)\\ \Rightarrow 4\mathrm{x}+\mathrm{y}=36\\ \text{⇒}\frac{x}{9}+\frac{y}{36}=1\Rightarrow \boxed{\alpha +\beta =45}\\ \end{array}\end{array}\end{array}\end{array}$