Let the function f x = 2 x 2 - log e x , x > 0 , be decreasing in 0 , a and increasing in a , 4 . A…

Given, $f\left(x\right)=2x^2-{\log }_{e}x$

$\Rightarrow f^{\text{'}}\left(x\right)=4x-\frac{1}{x}$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{4x^2-1}{x}$

$\Rightarrow f^{\text{'}}\left(x\right)=0\Rightarrow 4x^2-1=0\Rightarrow x=\pm \frac{1}{2}$

But given $x>0 \text{so} x=\frac{1}{2}$

So function is decreasing in $\left(0,\frac{1}{2}\right)$ and increasing in the interval $\left(\frac{1}{2},\infty \right)$

So, $a=\frac{1}{2}$

Now equation of parabola will be $y^2=2x$

Now tangent to $y^2=2x$ will be given by,

$y=mx+\frac{1}{2m}$, given this tangent passes through $\left(8a,8a-1\right)\equiv \left(4,3\right)$,

So $3=4m+\frac{1}{2m}$

$\Rightarrow m=\frac{1}{2} \text{or }\frac{1}{4}$

So equation of tangent are $y=\frac{x}{2}+1 \text{or }y=\frac{x}{4}+2$

But $y=\frac{x}{2}+1$ passes through $\left(-2,0\right)$ so rejected as given in the question.

Now equation of normal at $P$ will be,

$y=-4x-2\left(\frac{1}{2}\right)\left(-4\right)-\frac{1}{2}{\left(-4\right)}^3$ $\left\{\text{as slope of normal =}\frac{-1}{{\displaystyle \frac{1}{4}}}\text{=-4}\right\}$

$\Rightarrow y=-4x+4+32$

$\Rightarrow y+4x=36$

$\Rightarrow \frac{x}{9}+\frac{y}{36}=1$

So, $\alpha =9,\beta =36$

So, $\alpha +\beta =45$