Curve `b^(2)x^(2) + a^(2)y^(2) = a^(2)b^(2) and m^(2)x^(2)- y^(2)l^(2) = l^(2)m^(2)` intersect each other at right-angle if:

Options

A. `a^(2)+b^(2)=l^(2)+m^(2)`
B. `a^(2)-b^(2)=l^(2)-m^(2)`
C. `a^(2)-b^(2)=l^(2)+m^(2)`
D. `a^(2)+b^(2) = l^(2)-m^(2)`

Answer

To determine the condition under which the curves \( b^2 x^2 + a^2 y^2 = a^2 b^2 \) and \( m^2 x^2 - y^2 l^2 = l^2 m^2 \) intersect at right angles, we will follow these steps:

### Step 1: Rewrite the equations in standard form
1. **Ellipse Equation**:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
This is obtained by dividing the entire equation \( b^2 x^2 + a^2 y^2 = a^2 b^2 \) by \( a^2 b^2 \).

2. **Hyperbola Equation**:
\[ \frac{x^2}{l^2} - \frac{y^2}{m^2} = 1 \]
This is obtained by dividing the entire equation \( m^2 x^2 - y^2 l^2 = l^2 m^2 \) by \( l^2 m^2 \).

### Step 2: Differentiate both equations to find slopes
1. **Differentiate the ellipse**:
\[ \frac{d}{dx}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) = 0 \]
Using implicit differentiation:
\[ \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \]
Solving for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{b^2 x}{a^2 y} \]

2. **Differentiate the hyperbola**:
\[ \frac{d}{dx}\left(\frac{x^2}{l^2} - \frac{y^2}{m^2}\right) = 0 \]
Using implicit differentiation:
\[ \frac{2x}{l^2} - \frac{2y}{m^2} \frac{dy}{dx} = 0 \]
Solving for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{x m^2}{y l^2} \]

### Step 3: Set the product of slopes equal to -1
For the curves to intersect at right angles, the product of their slopes must equal -1:
\[ \left(-\frac{b^2 x}{a^2 y}\right) \left(\frac{x m^2}{y l^2}\right) = -1 \]
This simplifies to:
\[ \frac{b^2 m^2 x^2}{a^2 y^2 l^2} = 1 \]
Rearranging gives us:
\[ b^2 m^2 x^2 = a^2 y^2 l^2 \]

### Step 4: Use the equations to eliminate variables
From the equations of the ellipse and hyperbola, we can express \( y^2 \) in terms of \( x^2 \):
1. From the ellipse:
\[ y^2 = b^2 \left(1 - \frac{x^2}{a^2}\right) \]
2. Substitute \( y^2 \) into the condition:
\[ b^2 m^2 x^2 = a^2 \left(b^2 \left(1 - \frac{x^2}{a^2}\right)\right) l^2 \]
Simplifying this gives:
\[ b^2 m^2 x^2 = a^2 b^2 l^2 - a^2 b^2 \frac{x^2}{a^2} l^2 \]
This leads to:
\[ b^2 m^2 x^2 + b^2 l^2 x^2 = a^2 b^2 l^2 \]
Factoring out \( x^2 \):
\[ x^2 (b^2 m^2 + b^2 l^2) = a^2 b^2 l^2 \]
Thus:
\[ x^2 = \frac{a^2 l^2}{m^2 + l^2} \]

### Step 5: Substitute back to find the relationship
Substituting \( x^2 \) back into the equation for \( y^2 \):
\[ y^2 = b^2 \left(1 - \frac{a^2 l^2}{a^2(m^2 + l^2)}\right) \]
This leads to the final condition:
\[ m^2 + b^2 = a^2 - l^2 \]

### Final Condition
Thus, the curves intersect at right angles if:
\[ m^2 + b^2 = a^2 - l^2 \]