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

Options

A. 16
B. 12
C. 8
D. 4

Answer

To find 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 can follow these steps:

### Step 1: Differentiate the first curve
The first curve is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{12} = 1 \]
Differentiating both sides with respect to \( x \):
\[ \frac{2x}{a^2} + \frac{2y}{12} \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx} = -\frac{12x}{a^2y} \]
Let this slope be \( m_1 \).

### Step 2: Differentiate the second curve
The second curve is given by:
\[ y^3 = 8x \]
Differentiating both sides with respect to \( x \):
\[ 3y^2 \frac{dy}{dx} = 8 \]
Rearranging gives:
\[ \frac{dy}{dx} = \frac{8}{3y^2} \]
Let this slope be \( m_2 \).

### Step 3: Use the condition for perpendicularity
For the curves to intersect at right angles, the product of their slopes must equal -1:
\[ m_1 \cdot m_2 = -1 \]
Substituting the expressions for \( m_1 \) and \( m_2 \):
\[ \left(-\frac{12x_1}{a^2y_1}\right) \cdot \left(\frac{8}{3y_1^2}\right) = -1 \]
This simplifies to:
\[ \frac{96x_1}{3a^2y_1^3} = 1 \]
Thus, we have:
\[ \frac{96x_1}{3a^2y_1^3} = 1 \implies 96x_1 = 3a^2y_1^3 \]
So,
\[ a^2 = \frac{96x_1}{3y_1^3} = \frac{32x_1}{y_1^3} \]

### Step 4: Substitute \( y_1 \) from the second curve
From the second curve \( y^3 = 8x \), we can express \( x_1 \) in terms of \( y_1 \):
\[ x_1 = \frac{y_1^3}{8} \]
Substituting this into the equation for \( a^2 \):
\[ a^2 = \frac{32 \left(\frac{y_1^3}{8}\right)}{y_1^3} = \frac{32}{8} = 4 \]

### Conclusion
Thus, the value of \( a^2 \) is:
\[ \boxed{4} \]