If the curves ` (x ^(2))/(a ^(2))+ (y^(2))/(4)= 1 and y ^(2)= 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 of finding the value of \( a \) such that the curves

\[ \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \]

and

\[ y^2 = 16x \]

intersect at right angles, we will follow these steps:

### Step 1: Find the slopes of the tangents to both curves.

1. **Differentiate the first curve**:

Start with the equation of the first curve:

\[ \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \]

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

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

Simplifying gives:

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

Rearranging for \( \frac{dy}{dx} \):

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

Thus, the slope \( m_1 \) of the tangent to the first curve is:

\[ m_1 = -\frac{4x}{a^2 y} \]

2. **Differentiate the second curve**:

The second curve is given by:

\[ y^2 = 16x \]

Differentiate both sides:

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

Rearranging gives:

\[ \frac{dy}{dx} = \frac{8}{y} \]

Thus, the slope \( m_2 \) of the tangent to the second curve is:

\[ m_2 = \frac{8}{y} \]

### Step 2: Set up the condition for the curves to intersect at right angles.

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 \( m_2 \):

\[ \left(-\frac{4x}{a^2 y}\right) \cdot \left(\frac{8}{y}\right) = -1 \]

This simplifies to:

\[ -\frac{32x}{a^2 y^2} = -1 \]

Removing the negative signs gives:

\[ \frac{32x}{a^2 y^2} = 1 \]

### Step 3: Substitute \( y^2 \) from the second curve.

From the second curve, we know that:

\[ y^2 = 16x \]

Substituting this into the equation gives:

\[ \frac{32x}{a^2 (16x)} = 1 \]

This simplifies to:

\[ \frac{32}{16a^2} = 1 \]

### Step 4: Solve for \( a^2 \).

Cross-multiplying gives:

\[ 32 = 16a^2 \]

Dividing both sides by 16:

\[ a^2 = 2 \]

Taking the square root gives:

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

### Final Answer:

The value of \( a \) is \( \sqrt{2} \) or \( -\sqrt{2} \).

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