If the curves `(x^2)/(a^2)+(y^2)/(b^2)=1 and (x^2)/(l^2)-(y^2)/(m^2)=1`cut each other orthogonally then.....

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 solve the problem of determining the condition under which the curves \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (an ellipse) and \( \frac{x^2}{l^2} - \frac{y^2}{m^2} = 1 \) (a hyperbola) cut each other orthogonally, we can follow these steps:

### Step 1: Identify the curves
The first curve is an ellipse given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The second curve is a hyperbola given by:
\[ \frac{x^2}{l^2} - \frac{y^2}{m^2} = 1 \]

### Step 2: Find the foci of the curves
For the ellipse, the foci are located at:
\[ (\pm c, 0) \quad \text{where } c = \sqrt{a^2 - b^2} \]
For the hyperbola, the foci are located at:
\[ (\pm C, 0) \quad \text{where } C = \sqrt{l^2 + m^2} \]

### Step 3: Set the foci equal for orthogonality
For the two curves to intersect orthogonally, their foci must coincide. Thus, we have:
\[ \sqrt{a^2 - b^2} = \sqrt{l^2 + m^2} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ a^2 - b^2 = l^2 + m^2 \]

### Conclusion
Thus, the condition for the curves to cut each other orthogonally is:
\[ a^2 - b^2 = l^2 + m^2 \]