If the curves `(x^2)/(a^2)+(y^2)/4=1` and `y^3=16x` intersects at right angles then `a^2`=

Options

A. `1/2`
B. `3/4`
C. `4/3`
D. none

Answer

To solve the problem of finding \( a^2 \) such that the curves \( \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \) and \( y^3 = 16x \) intersect at right angles, we can follow these steps:

### Step 1: Identify the curves
The first curve is an ellipse given by:
\[ C_1: \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \]
The second curve is a cubic curve given by:
\[ C_2: y^3 = 16x \]

### Step 2: Find the intersection points
Let the intersection point be \( (h, k) \). We need to satisfy both equations at this point. From the second curve, we have:
\[ k^3 = 16h \quad \text{(1)} \]

### Step 3: Find the derivatives (slopes of tangents)
To find the slopes of the tangents at the intersection point, we need to differentiate both curves.

**For the first curve \( C_1 \)**:
Using implicit differentiation:
\[ \frac{d}{dx}\left(\frac{x^2}{a^2} + \frac{y^2}{4}\right) = 0 \]
This gives:
\[ \frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \]
Rearranging, we find:
\[ \frac{dy}{dx} = -\frac{2x}{a^2} \cdot \frac{4}{2y} = -\frac{4x}{a^2y} \]
At the point \( (h, k) \):
\[ M_1 = -\frac{4h}{a^2k} \quad \text{(2)} \]

**For the second curve \( C_2 \)**:
Differentiating gives:
\[ 3y^2 \frac{dy}{dx} = 16 \]
Thus,
\[ \frac{dy}{dx} = \frac{16}{3y^2} \]
At the point \( (h, k) \):
\[ M_2 = \frac{16}{3k^2} \quad \text{(3)} \]

### Step 4: Set up the condition for perpendicular tangents
Since the curves intersect at right angles, the product of their slopes must equal -1:
\[ M_1 \cdot M_2 = -1 \]
Substituting the values from (2) and (3):
\[ \left(-\frac{4h}{a^2k}\right) \cdot \left(\frac{16}{3k^2}\right) = -1 \]
This simplifies to:
\[ -\frac{64h}{3a^2k^3} = -1 \]
Thus,
\[ \frac{64h}{3a^2k^3} = 1 \quad \Rightarrow \quad 64h = 3a^2k^3 \quad \text{(4)} \]

### Step 5: Substitute equation (1) into equation (4)
From equation (1), we have \( k^3 = 16h \). Substitute this into equation (4):
\[ 64h = 3a^2(16h) \]
This simplifies to:
\[ 64h = 48a^2h \]
Assuming \( h \neq 0 \), we can divide both sides by \( h \):
\[ 64 = 48a^2 \]
Solving for \( a^2 \):
\[ a^2 = \frac{64}{48} = \frac{4}{3} \]

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