Let the function f(x) = 2 x ^ 2 - _ e x, x > 0 , be decreasing in...

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

$⇒ {f}^{'}\left(x\right) = 4x - \dfrac{1}{x}$

$⇒ {f}^{'}\left(x\right) = \dfrac{4{x}^{2} - 1}{x}$

$⇒ {f}^{'}\left(x\right) = 0 ⇒ 4{x}^{2} - 1 = 0 ⇒ x = \pm \dfrac{1}{2}$

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

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

So, $a = \dfrac{1}{2}$

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

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

$y = mx + \dfrac{1}{2m}$, given this tangent passes through \left(8a, 8a - 1\right) ≡ \left(4, 3\right)$,</p><p>So$3 = 4m + \dfrac{1}{2m}$</p><p>$⇒ m = \dfrac{1}{2}\text{or}\dfrac{1}{4}$</p><p>So equation of tangent are$y = \dfrac{x}{2} + 1\text{or}y = \dfrac{x}{4} + 2$</p><p>But$y = \dfrac{x}{2} + 1$passes through \left(- 2, 0\right)$ so rejected as given in the question.

Now equation of normal at $P$ will be,

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

$⇒ y = - 4x + 4 + 32$

$⇒ y + 4x = 36$

$⇒ \dfrac{x}{9} + \dfrac{y}{36} = 1$

So, $\alpha = 9, \beta = 36$

So, $\alpha + \beta = 45$