If the curves `x^(2)/a^(2)+ y^(2)/4 = 1 ` and `y^(3) = 16x` intersect at right angles, then `a^(2)` is equal to

Options

A. `1/3`
B. `1/4`
C. `4/3`
D. none of these

Answer

To solve the problem, we need to find the value of \( a^2 \) such that the curves \( \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \) and \( y^3 = 16x \) intersect at right angles.

### Step 1: Differentiate the first curve
The first curve is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \]
To find the slope of the tangent line, we differentiate this implicitly with respect to \( x \):
\[ \frac{d}{dx}\left(\frac{x^2}{a^2}\right) + \frac{d}{dx}\left(\frac{y^2}{4}\right) = 0 \]
This gives:
\[ \frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \]
Simplifying, we have:
\[ \frac{2x}{a^2} + \frac{y}{2} \frac{dy}{dx} = 0 \]
Rearranging for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{4x}{a^2y} \]

### Step 2: Differentiate the second curve
The second curve is given by:
\[ y^3 = 16x \]
Differentiating implicitly with respect to \( x \):
\[ 3y^2 \frac{dy}{dx} = 16 \]
So,
\[ \frac{dy}{dx} = \frac{16}{3y^2} \]

### Step 3: Set up the condition for perpendicular tangents
The curves intersect at right angles, which means the product of their slopes at the point of intersection must equal -1:
\[ \left(-\frac{4x_1}{a^2 y_1}\right) \left(\frac{16}{3y_1^2}\right) = -1 \]
This simplifies to:
\[ \frac{64x_1}{3a^2y_1^3} = 1 \]
Thus,
\[ 64x_1 = 3a^2y_1^3 \]
From this, we can express \( a^2 \):
\[ a^2 = \frac{64x_1}{3y_1^3} \]

### Step 4: Find the intersection point
To find \( x_1 \) and \( y_1 \), we substitute \( y_1 \) from the second curve into the first curve. From \( y^3 = 16x \), we have:
\[ x = \frac{y^3}{16} \]
Substituting into the first curve:
\[ \frac{\left(\frac{y^3}{16}\right)^2}{a^2} + \frac{y^2}{4} = 1 \]
This simplifies to:
\[ \frac{y^6}{256a^2} + \frac{y^2}{4} = 1 \]
Multiplying through by \( 256a^2 \):
\[ y^6 + 64a^2y^2 = 256a^2 \]
Rearranging gives us:
\[ y^6 + 64a^2y^2 - 256a^2 = 0 \]

### Step 5: Solve for \( y_1 \)
Let \( z = y^2 \). The equation becomes:
\[ z^3 + 64a^2z - 256a^2 = 0 \]
Using the Rational Root Theorem or synthetic division, we can find potential rational roots. Testing \( z = 4 \):
\[ 4^3 + 64a^2(4) - 256a^2 = 0 \]
This simplifies to:
\[ 64 + 256a^2 - 256a^2 = 0 \]
Thus, \( y^2 = 4 \) or \( y_1 = 2 \).

### Step 6: Substitute back to find \( x_1 \)
Substituting \( y_1 = 2 \) into \( y^3 = 16x \):
\[ 2^3 = 16x \implies 8 = 16x \implies x_1 = \frac{1}{2} \]

### Step 7: Calculate \( a^2 \)
Now substituting \( x_1 \) and \( y_1 \) back into the equation for \( a^2 \):
\[ a^2 = \frac{64 \cdot \frac{1}{2}}{3 \cdot 2^3} = \frac{32}{3 \cdot 8} = \frac{32}{24} = \frac{4}{3} \]

Thus, the value of \( a^2 \) is:
\[ \boxed{\frac{4}{3}} \]