If `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1(agtb) and x^(2)-y^(2)=c^(2)` cut at right angles then

Options

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

Answer

To solve the problem, we need to find the relationship between \( a^2 \), \( b^2 \), and \( c^2 \) given the equations of a hyperbola and a rectangular hyperbola that intersect at right angles.

### Step-by-Step Solution:

1. **Identify the Equations**:
The first equation is the equation of an ellipse:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b) \]
The second equation is the equation of a hyperbola:
\[ x^2 - y^2 = c^2 \]

2. **Differentiate the First Equation**:
Differentiate the first equation with respect to \( x \):
\[ \frac{d}{dx}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) = 0 \]
This gives:
\[ \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \]
Rearranging for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{y b^2}{x a^2} \]

3. **Differentiate the Second Equation**:
Differentiate the second equation with respect to \( x \):
\[ \frac{d}{dx}(x^2 - y^2) = 0 \]
This gives:
\[ 2x - 2y \frac{dy}{dx} = 0 \]
Rearranging for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{x}{y} \]

4. **Set the Slopes to be Negative Reciprocal**:
Since the curves cut at right angles, we have:
\[ \left(-\frac{y b^2}{x a^2}\right) \cdot \left(\frac{x}{y}\right) = -1 \]
Simplifying this gives:
\[ -\frac{b^2}{a^2} = -1 \implies \frac{b^2}{a^2} = 1 \]

5. **Cross Multiply**:
Cross multiplying gives:
\[ b^2 = a^2 \]
However, since \( a > b \), we need to find a different relationship.

6. **Substituting Values**:
From the hyperbola equation \( x^2 - y^2 = c^2 \), substitute \( x^2 \) and \( y^2 \) using the previous results:
Let \( x^2 = \frac{a^2}{2} \) and \( y^2 = \frac{b^2}{2} \):
\[ \frac{a^2}{2} - \frac{b^2}{2} = c^2 \]
This simplifies to:
\[ a^2 - b^2 = 2c^2 \]

### Final Result:
Thus, the relationship we derived is:
\[ a^2 - b^2 = 2c^2 \]