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

Options

A. `a=pm1`
B. `a= pm sqrt3`
C. ` a= pm sqrt3`
D. `a = pm sqrt2`

Answer

To solve the problem, we need to find the value of \( a \) such that the curves

1. \( \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \) (an ellipse)
2. \( y^3 = 16x \) (a cubic curve)

intersect at right angles.

### Step 1: Find the slopes of both curves at the point of intersection

Let the point of intersection be \( P(h, k) \).

**For the first curve:**

Differentiate \( \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \):

\[ \frac{d}{dx}\left(\frac{x^2}{a^2}\right) + \frac{d}{dx}\left(\frac{y^2}{4}\right) = 0 \]

Using implicit differentiation:

\[ \frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0 \]

This simplifies to:

\[ \frac{2y}{4} \frac{dy}{dx} = -\frac{2x}{a^2} \]

Thus,

\[ \frac{dy}{dx} = -\frac{4x}{a^2y} \quad \text{(let's call this } m_1\text{)} \]

**For the second curve:**

Differentiate \( y^3 = 16x \):

\[ 3y^2 \frac{dy}{dx} = 16 \]

Thus,

\[ \frac{dy}{dx} = \frac{16}{3y^2} \quad \text{(let's call this } m_2\text{)} \]

### Step 2: Evaluate the slopes at the point of intersection

At point \( P(h, k) \):

\[ m_1 = -\frac{4h}{a^2 k} \]
\[ m_2 = \frac{16}{3k^2} \]

### Step 3: Set up the condition for perpendicularity

Since the curves intersect at right angles, we have:

\[ m_1 \cdot m_2 = -1 \]

Substituting the values of \( m_1 \) and \( m_2 \):

\[ \left(-\frac{4h}{a^2 k}\right) \cdot \left(\frac{16}{3k^2}\right) = -1 \]

This simplifies to:

\[ \frac{64h}{3a^2 k^3} = 1 \]

### Step 4: Express \( h \) in terms of \( k \)

Since point \( P(h, k) \) lies on the second curve \( y^3 = 16x \):

\[ k^3 = 16h \implies h = \frac{k^3}{16} \]

### Step 5: Substitute \( h \) into the perpendicularity condition

Substituting \( h \) in the equation:

\[ \frac{64 \left(\frac{k^3}{16}\right)}{3a^2 k^3} = 1 \]

This simplifies to:

\[ \frac{4}{3a^2} = 1 \implies 3a^2 = 4 \implies a^2 = \frac{4}{3} \]

### Step 6: Find the value of \( a \)

Taking the square root:

\[ a = \pm \frac{2}{\sqrt{3}} \]

### Final Answer

Thus, the required value of \( a \) is:

\[ a = \pm \frac{2}{\sqrt{3}} \]

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