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Step-by-Step Solution
Step 1: Find the value of “a” using the function’s monotonicity
We are given the function
f(x) = 2x^2 - \log_e x \quad (x>0).
Its derivative is
f'(x) = \frac{d}{dx}(2x^2) - \frac{d}{dx}(\log_e x) = 4x - \frac{1}{x} = \frac{4x^2 - 1}{x}.
For f(x) to be decreasing, we need f'(x) < 0 , and for it to be increasing, we need f'(x) > 0 . The critical point occurs when f'(x) = 0 , i.e.
4x^2 - 1 = 0 \quad \Longrightarrow \quad x^2 = \frac{1}{4} \quad \Longrightarrow \quad x = \frac{1}{2}.
Thus, f(x) is decreasing on (0,\tfrac12) and increasing on (\tfrac12, \infty) . Hence we deduce
a = \frac{1}{2}.
Step 2: Identify the parabola and point through which the tangent passes
With a = \tfrac12 , the parabola
y^2 = 4ax
becomes
y^2 = 2x.
We are told a tangent to this parabola passes through the point
(8a, \, 8a - 1).
Substituting a = \tfrac12 gives the point
(4, \, 3).
Step 3: Write the general form of the tangent and find its slope
A standard form of the tangent to
y^2 = 2x
is
y = mx + \frac{1}{2m},
where m is the slope of the tangent. Since it passes through (4, 3) , we substitute x = 4 and y = 3 :
3 = m \cdot 4 + \frac{1}{2m}
\quad\Longrightarrow\quad
3 = 4m + \frac{1}{2m}.
Multiplying through by 2m , we get
6m = 8m^2 + 1 \quad\Longrightarrow\quad 8m^2 - 6m + 1 = 0.
Solving for m :
8m^2 - 6m + 1 = 0.
You can solve this quadratic via the quadratic formula or by factoring if possible. Let us suppose the solutions found are
m_1, m_2.
(Indeed, from the provided outlines, the solutions are m = \tfrac12 or m = \tfrac14 .)
Step 4: Determine which slope is valid by checking the second point
The second condition is that the tangent does not pass through
\Bigl(-\frac{1}{a},\,0\Bigr) = \Bigl(-2,\, 0\Bigr).
We check each candidate slope:
For m = \tfrac12 , the tangent is
y = \tfrac12\,x + 1.
Substitute x=-2 :
y = \tfrac12(-2) + 1 = -1 + 1 = 0.
So it does pass through (-2, 0) . This violates the condition, so we discard m = \tfrac12 .
For m = \tfrac14 , the tangent is
y = \tfrac14\,x + 2.
Substitute x = -2 :
y = \tfrac14 \,(-2) + 2 = -\tfrac12 + 2 = \tfrac{3}{2},
which is not 0 . Hence it does not pass through (-2, 0) , so this slope is valid.
Step 5: Find the coordinates of the point of tangency “P”
We now know the slope of the tangent is m = \tfrac14 . For the parabola
y^2 = 2x,
the slope at any point (x, y) can also be found by implicit differentiation:
2y \,\frac{dy}{dx} = 2 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1}{y}.
Setting this equal to \tfrac14 :
\frac{1}{y} = \tfrac14
\quad\Longrightarrow\quad
y = 4.
Since y^2 = 2x , we get
16 = 2x \quad\Longrightarrow\quad x = 8.
Thus, the point of tangency is
P = (8,\,4).
Step 6: Write the equation of the normal at P
The slope of the normal is the negative reciprocal of the slope of the tangent. Since the tangent’s slope is m = \tfrac14 , the normal’s slope is
-4.
Hence, the normal line at (8, 4) is
y - 4 = -4\,(x - 8).
Simplify:
y - 4 = -4x + 32
\quad\Longrightarrow\quad
y = -4x + 36.
Step 7: Convert the normal to intercept form and find \alpha + \beta
In the intercept form
\frac{x}{\alpha} + \frac{y}{\beta} = 1,
we rearrange
y = -4x + 36
to
4x + y = 36.
Divide throughout by 36,
\frac{4x}{36} + \frac{y}{36} = 1
\quad\Longrightarrow\quad
\frac{x}{9} + \frac{y}{36} = 1.
So we identify
\alpha = 9 \quad\text{and}\quad \beta = 36.
Therefore,
\alpha + \beta = 9 + 36 = 45.
Final Answer
The required sum of \alpha + \beta is
45.
