Key Concepts and Formulas

• Monotonicity of a function: A function $f(x)$ is decreasing when $f'(x) < 0$ and increasing when $f'(x) > 0$. The point where $f'(x) = 0$ is a critical point, which can be a local minimum or maximum.
• Equation of a tangent to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $ty = x + at^2$.
• Equation of a normal to the parabola $y^2 = 4ax$ at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$.

Step-by-Step Solution

Step 1: Find the derivative of f(x) and determine the critical point.

We are given $f(x) = 2x^2 - \log_e x$. To find where $f(x)$ is decreasing or increasing, we need to analyze its derivative $f'(x)$. $f'(x) = \frac{d}{dx}(2x^2 - \log_e x) = 4x - \frac{1}{x}$ The function is decreasing in $(0, a)$ and increasing in $(a, 4)$. Therefore, $f'(x) = 0$ at $x = a$. $4a - \frac{1}{a} = 0$ $4a^2 = 1$ $a^2 = \frac{1}{4}$ Since $x > 0$, we have $a = \frac{1}{2}$.

Step 2: Find the equation of the tangent to the parabola.

We have the parabola $y^2 = 4ax$, and we know $a = \frac{1}{2}$. So, the parabola is $y^2 = 4(\frac{1}{2})x = 2x$. Let $P$ be the point $(at^2, 2at)$ on the parabola. Since $a = \frac{1}{2}$, the point $P$ is $(\frac{1}{2}t^2, t)$. The equation of the tangent to the parabola $y^2 = 4ax$ at $(at^2, 2at)$ is $ty = x + at^2$. Substituting $a = \frac{1}{2}$, we get $ty = x + \frac{1}{2}t^2$ The tangent passes through the point $(8a, 8a - 1)$, which is $(8(\frac{1}{2}), 8(\frac{1}{2}) - 1) = (4, 3)$. Substituting this into the tangent equation: $3t = 4 + \frac{1}{2}t^2$ $6t = 8 + t^2$ $t^2 - 6t + 8 = 0$ $(t - 2)(t - 4) = 0$ So, $t = 2$ or $t = 4$.

Step 3: Check the condition that the tangent does not pass through (-1/a, 0).

The tangent equation is $ty = x + \frac{1}{2}t^2$. We are given that the tangent does not pass through $(-\frac{1}{a}, 0) = (-2, 0)$. Substituting $(-2, 0)$ into the tangent equation: $t(0) = -2 + \frac{1}{2}t^2$ $0 = -2 + \frac{1}{2}t^2$ $t^2 = 4$ $t = \pm 2$ Since the tangent does not pass through $(-2, 0)$, and $t=2$ would make it pass through $(-2,0)$, we must have $t \ne 2$. So, $t = 4$.

Step 4: Find the equation of the normal.

The equation of the normal to the parabola $y^2 = 4ax$ at $(at^2, 2at)$ is $y = -tx + 2at + at^3$. With $a = \frac{1}{2}$ and $t = 4$, we have: $y = -4x + 2(\frac{1}{2})(4) + (\frac{1}{2})(4^3)$ $y = -4x + 4 + 32$ $y = -4x + 36$ $4x + y = 36$ Divide by 36 to get the form $\frac{x}{\alpha} + \frac{y}{\beta} = 1$: $\frac{4x}{36} + \frac{y}{36} = 1$ $\frac{x}{9} + \frac{y}{36} = 1$ So, $\alpha = 9$ and $\beta = 36$.

Step 5: Calculate α + β.

We need to find $\alpha + \beta = 9 + 36 = 45$. However, the answer should be 4. There is an error. Let's go back to Step 3. When $t=2$, the tangent equation is $2y=x+2$. It passes through $(4,3)$ since $2(3)=4+2$. It also passes through $(-2,0)$ since $2(0)=-2+2$. When $t=4$, the tangent equation is $4y=x+8$. It passes through $(4,3)$ since $4(3)=4+8$. It does NOT pass through $(-2,0)$ since $4(0) \ne -2+8$. Hence, $t=4$. The equation of the normal is $y = -tx + 2at + at^3 = -4x + 2(\frac{1}{2})(4) + (\frac{1}{2})(4)^3 = -4x + 4 + 32 = -4x + 36$. Then $4x+y=36 \implies \frac{x}{9} + \frac{y}{36}=1$. Hence $\alpha=9, \beta=36$, so $\alpha+\beta=45$.

Let's rethink this. If the equation of the normal is $\frac{x}{\alpha}+\frac{y}{\beta}=1$, then we want to find $\alpha+\beta$. Since $t=4$ and $a=\frac{1}{2}$, the point P is $(\frac{1}{2}(4^2), 4) = (8,4)$. The slope of the tangent is $t = 4$. Therefore, the slope of the normal is $-\frac{1}{4}$. The equation of the normal is $y-4 = -\frac{1}{4}(x-8)$ or $4y-16 = -x+8$ or $x+4y=24$. Then $\frac{x}{24} + \frac{y}{6} = 1$, so $\alpha=24$ and $\beta=6$. Then $\alpha+\beta=30$.

Let's re-derive the normal equation $y=-tx + 2at + at^3$. We want the normal at $(at^2, 2at) = (8,4)$ with $t=4$ and $a=\frac{1}{2}$. The slope of the tangent is $t=4$ so slope of normal is $-\frac{1}{t}=-\frac{1}{4}$. $y-2at = -\frac{1}{t}(x-at^2) \implies y-4 = -\frac{1}{4}(x-8) \implies 4y-16=-x+8 \implies x+4y=24 \implies \frac{x}{24}+\frac{y}{6}=1$. $\alpha=24, \beta=6$. $\alpha+\beta=30$.

Let's try $t=-2$ (even though the tangent does pass through $(-2,0)$). Then the normal is $y = 2x + 2(\frac{1}{2})(-2) + (\frac{1}{2})(-2)^3 = 2x - 2 - 4 = 2x - 6$ or $-2x+y=-6$ or $2x-y=6$ or $\frac{x}{3} + \frac{y}{-6}=1$. Then $\alpha=3, \beta=-6$, so $\alpha+\beta=-3$.

The correct answer given is 4. There must be an error in the statement of the problem. Let's assume that the equation of the normal at P is $\alpha x + \beta y = 1$. With $t=4$, we have $x+4y=24$, so $\frac{1}{24} x + \frac{4}{24}y=1$ or $\frac{1}{24} x + \frac{1}{6} y = 1$. Thus $\alpha=\frac{1}{24}$ and $\beta=\frac{1}{6}$. Thus $\frac{1}{\alpha} + \frac{1}{\beta} = 24+6=30$.

The problem states: If the equation of the normal at $P$ is : $\frac{x}{\alpha}+\frac{y}{\beta}=1$, then $\alpha+\beta$ is equal to ________________.

$x+4y=24$ implies $\frac{x}{24} + \frac{y}{6} = 1$, so $\alpha=24$ and $\beta=6$, then $\alpha+\beta=30$.

If the equation is $\alpha x + \beta y = 1$, then $\alpha = \frac{1}{24}$ and $\beta = \frac{1}{6}$, so $\frac{1}{\alpha} + \frac{1}{\beta} = 24+6 = 30$.

If the equation of the normal is given as $\frac{x}{\alpha} + \frac{y}{\beta} = 1$, then $\alpha = 24$ and $\beta = 6$. Then $\alpha+\beta = 24+6 = 30$.

Let's consider a scenario where the normal equation is $Ax+By=C$. Dividing by $C$, we get $\frac{A}{C}x + \frac{B}{C}y=1$. Then $\alpha = \frac{C}{A}$ and $\beta=\frac{C}{B}$. We have $x+4y=24$, so $A=1, B=4, C=24$. Then $\alpha=24, \beta=6$ and $\alpha+\beta=30$.

If we want $\alpha + \beta = 4$, then we need to check the original problem. It seems there's an error.

Common Mistakes & Tips

• Double-check the signs when calculating the derivative and the equation of the normal.
• Be careful when solving quadratic equations and consider all possible solutions.
• Remember that the tangent does not pass through a given point, so you must exclude the solution where it does.

Summary

We analyzed the function to find the value of $a$, then found the equation of the tangent to the parabola and used the given condition to find the value of $t$. Using this, we derived the equation of the normal and determined the values of $\alpha$ and $\beta$. The sum $\alpha + \beta$ is 30. There appears to be an error in the provided correct answer.

Final Answer

The final answer is \boxed{30}.