If the curves `(x ^(2))/(a ^(2)) + (y ^(2))/(12) =1 and y ^(3) =8x` intersect at right angles, then the value of ` a ^(2)` is equal to

Answer

To solve the problem of finding the value of \( a^2 \) such that the curves

\[ \frac{x^2}{a^2} + \frac{y^2}{12} = 1 \]

and

\[ y^3 = 8x \]

intersect at right angles, we will follow these steps:

### Step 1: Differentiate the first curve

We start with the first equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{12} = 1 \]

Differentiating both sides with respect to \( x \):

\[ \frac{d}{dx}\left(\frac{x^2}{a^2}\right) + \frac{d}{dx}\left(\frac{y^2}{12}\right) = 0 \]

Using the chain rule, we get:

\[ \frac{2x}{a^2} + \frac{2y}{12} \frac{dy}{dx} = 0 \]

### Step 2: Solve for \( \frac{dy}{dx} \)

Rearranging the equation to isolate \( \frac{dy}{dx} \):

\[ \frac{2y}{12} \frac{dy}{dx} = -\frac{2x}{a^2} \]

This simplifies to:

\[ \frac{dy}{dx} = -\frac{12x}{a^2 y} \]

### Step 3: Differentiate the second curve

Now we differentiate the second equation:

\[ y^3 = 8x \]

Differentiating both sides with respect to \( x \):

\[ 3y^2 \frac{dy}{dx} = 8 \]

### Step 4: Solve for \( \frac{dy}{dx} \) for the second curve

Rearranging gives:

\[ \frac{dy}{dx} = \frac{8}{3y^2} \]

### Step 5: Set the slopes equal to the condition for perpendicularity

Since the curves intersect at right angles, the product of their slopes must equal -1:

\[ \left(-\frac{12x}{a^2 y}\right) \left(\frac{8}{3y^2}\right) = -1 \]

### Step 6: Simplify the equation

Multiplying the left-hand side:

\[ -\frac{96x}{3a^2 y^3} = -1 \]

This simplifies to:

\[ \frac{96x}{3a^2 y^3} = 1 \]

### Step 7: Solve for \( a^2 \)

Rearranging gives:

\[ a^2 = \frac{96x}{3y^3} \]

### Step 8: Substitute \( y^3 \) from the second curve

From the second curve \( y^3 = 8x \), we substitute \( y^3 \):

\[ a^2 = \frac{96x}{3(8x)} = \frac{96x}{24x} = 4 \]

### Final Answer

Thus, the value of \( a^2 \) is:

\[ \boxed{4} \]