The value(s) of a for which two curves `y=ax^(2)+ax+(1)/(24)andx=ay^(2)+ay+(1)/(24)` touch each other is/are

Options

A. `(2)/(3)`
B. `(1)/(3)`
C. `(3)/(2)`
D. `(1)/(2)`

Explanation

To find the values of \( a \) for which the curves \( y = ax^2 + ax + \frac{1}{24} \) and \( x = ay^2 + ay + \frac{1}{24} \) touch each other, we will follow these steps:

### Step 1: Set the equations equal to each other
Since the curves are symmetric about the line \( y = x \), we can set \( y = x \) in the first equation:
\[ x = ax^2 + ax + \frac{1}{24} \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ ax^2 + (a - 1)x + \frac{1}{24} = 0 \]

### Step 3: Apply the condition for tangency
For the two curves to touch, the quadratic equation must have exactly one solution. This occurs when the discriminant is zero. The discriminant \( D \) for the quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:
\[ D = B^2 - 4AC \]
Here, \( A = a \), \( B = a - 1 \), and \( C = \frac{1}{24} \).

### Step 4: Calculate the discriminant
Setting the discriminant to zero:
\[ D = (a - 1)^2 - 4 \cdot a \cdot \frac{1}{24} = 0 \]
Expanding this gives:
\[ (a - 1)^2 - \frac{a}{6} = 0 \]

### Step 5: Simplify the equation
Expanding \( (a - 1)^2 \):
\[ a^2 - 2a + 1 - \frac{a}{6} = 0 \]
To eliminate the fraction, multiply the entire equation by 6:
\[ 6a^2 - 12a + 6 - a = 0 \]
This simplifies to:
\[ 6a^2 - 13a + 6 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 6 \), \( b = -13 \), and \( c = 6 \):
\[ a = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 6 \cdot 6}}{2 \cdot 6} \]
Calculating the discriminant:
\[ (-13)^2 = 169 \]
\[ 4 \cdot 6 \cdot 6 = 144 \]
Thus,
\[ 169 - 144 = 25 \]
So,
\[ a = \frac{13 \pm 5}{12} \]

### Step 7: Find the values of \( a \)
Calculating the two possible values:
1. \( a = \frac{18}{12} = \frac{3}{2} \)
2. \( a = \frac{8}{12} = \frac{2}{3} \)

### Final Answer
The values of \( a \) for which the two curves touch each other are:
\[ a = \frac{3}{2} \quad \text{and} \quad a = \frac{2}{3} \]

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