If the curves `2x^(2)+3y^(2)=6 and ax^(2)+4y^(2)=4` intersect orthogonally, then a =

Options

A. 2
B. 1
C. 3
D. none of these

Explanation

To solve the problem of finding the value of \( a \) such that the curves \( 2x^2 + 3y^2 = 6 \) and \( ax^2 + 4y^2 = 4 \) intersect orthogonally, we can follow these steps:

### Step 1: Rewrite the equations in standard form
The first curve can be rewritten as:
\[ \frac{x^2}{3} + \frac{y^2}{2} = 1 \]
This indicates that \( a^2 = 3 \) and \( b^2 = 2 \).

The second curve can be rewritten as:
\[ \frac{ax^2}{4} + \frac{y^2}{1} = 1 \]
This indicates that \( c^2 = \frac{4}{a} \) and \( d^2 = 1 \).

### Step 2: Use the condition for orthogonal intersection
For two curves to intersect orthogonally, the following condition must hold:
\[ a^2 - b^2 = c^2 - d^2 \]
Substituting the values we found:
\[ 3 - 2 = \frac{4}{a} - 1 \]

### Step 3: Simplify the equation
This simplifies to:
\[ 1 = \frac{4}{a} - 1 \]
Adding 1 to both sides gives:
\[ 2 = \frac{4}{a} \]

### Step 4: Solve for \( a \)
To find \( a \), we can cross-multiply:
\[ 2a = 4 \]
Dividing both sides by 2 gives:
\[ a = 2 \]

### Conclusion
Thus, the value of \( a \) for which the curves intersect orthogonally is:
\[ \boxed{2} \]