The values of $a$ for which $y = ax^2 + ax + \frac{1}{24}$, $x = ay^2 + ay + \frac{1}{24}$ touch each other is/are
Step-by-Step Solution
Key Concept: Two curves touch when they intersect at a point with the same tangent line. For the parabolas y = ax² + ax + 1/24 and x = ay² + ay + 1/24 (symmetric about y = x), contact points lie on y = x, requiring both the curve equation and equal derivatives dy/dx to be satisfied simultaneously at the point of tangency.
The point of contact lies on the line $y = x$ at $(a, a)$. The tangent slope is $\pm 1$ and relates to the curve by $a = a\sigma^2 + a\sigma + \frac{1}{24}$. Eliminating $\sigma$ from the tangent slope condition yields $\left(\frac{\pm 1 - \sigma}{2a}\right)^2 = a\left(\frac{\pm 1 - \sigma}{2a}\right) + a + \frac{1}{24}$. Solving gives $a = \frac{2}{3}, \frac{3}{2}$ and corresponding $\sigma = \frac{1}{3}, \frac{13 \pm \sqrt{601}}{12}$.
Correct Answer: 1,2,3,4