Let the function f(x)=2x^2- log_ex,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…

Given, $f(x)=2{x}^{2}-{\mathrm{log}}_{e}x$f(x)=2x2−loge​x

$$\Rightarrow {f}^{'}(x)=4x-\frac{1}{x}$$⇒f′(x)=4x−x1​

$$\Rightarrow {f}^{'}(x)=\frac{4{x}^{2}-1}{x}$$⇒f′(x)=x4x2−1​

$$\Rightarrow {f}^{'}(x)=0\Rightarrow 4{x}^{2}-1=0\Rightarrow x=\pm \frac{1}{2}$$⇒f′(x)=0⇒4x2−1=0⇒x=±21​

But given $x>0\text{so}x=\frac{1}{2}$x>0sox=21​

So function is decreasing in $(0,\frac{1}{2})$(0,21​) and increasing in the interval $(\frac{1}{2},\infty )$(21​,∞)

So, $a=\frac{1}{2}$a=21​

Now equation of parabola will be ${y}^{2}=2x$y2=2x

Now tangent to ${y}^{2}=2x$y2=2x will be given by,

$y=mx+\frac{1}{2m}$y=mx+2m1​, given this tangent passes through $(8a,8a-1)\equiv (4,3)$(8a,8a−1)≡(4,3),

So $3=4m+\frac{1}{2m}$3=4m+2m1​

$$\Rightarrow m=\frac{1}{2}\text{or }\frac{1}{4}$$⇒m=21​or 41​

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

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

Now equation of normal at $P$P will be,

$y=-4x-2(\frac{1}{2})(-4)-\frac{1}{2}{(-4)}^{3}$y=−4x−2(21​)(−4)−21​(−4)3 ${\text{as slope of normal =}\frac{-1}{\frac{1}{4}}\text{=-4}}$as slope of normal =41​−1​=-4

$$\Rightarrow y=-4x+4+32$$⇒y=−4x+4+32

$$\Rightarrow y+4x=36$$⇒y+4x=36

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

So, $\alpha =9,\beta =36$α=9,β=36

So, $\alpha +\beta =45$α+β=45