Question

If \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (\( a > b \)) and \( x^2 - y^2 = c^2 \) cut at right angles, then

• A. \( a^2 + b^2 = 2c^2 \)
• B. \( b^2 - a^2 = 2c^2 \)
• C. \( a^2 - b^2 = 2c^2 \)
• D. \( a^2 - b^2 = c^2 \) (Typo in Q paper)

Hint:

The condition for two curves, an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) and a confocal hyperbola \( \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 \), to be orthogonal is that they must be confocal. For a general ellipse and hyperbola as given, the orthogonality condition is derived by setting the product of the slopes of their tangents to -1 at the intersection point. This leads to the relation \( a^2 - b^2 = 2c^2 \).

Answer 1

The Correct Option is C

Two curves cut at right angles if their tangents at the point of intersection are perpendicular. This means the product of their slopes is -1.
Let the point of intersection be \( (x_0, y_0) \).

Step 1: Find the slopes of the tangents. For the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), differentiate with respect to x:
\[ \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = m_1 = -\frac{b^2x}{a^2y} \] At \( (x_0, y_0) \), the slope is \( m_1 = -\frac{b^2x_0}{a^2y_0} \).
For the hyperbola \( x^2 - y^2 = c^2 \), differentiate with respect to x: \[ 2x - 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = m_2 = \frac{x}{y} \] At \( (x_0, y_0) \), the slope is \( m_2 = \frac{x_0}{y_0} \).

Step 2: Apply the orthogonality condition \( m_1 m_2 = -1 \). \[ \left(-\frac{b^2x_0}{a^2y_0}\right) \left(\frac{x_0}{y_0}\right) = -1 \] \[ \frac{b^2x_0^2}{a^2y_0^2} = 1 \implies b^2x_0^2 = a^2y_0^2 \implies \frac{x_0^2}{a^2} = \frac{y_0^2}{b^2} \]

Step 3: Use the equations of the curves. Since \( (x_0, y_0) \) lies on the ellipse:
\[ \frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1 \] Substitute \( \frac{y_0^2}{b^2} = \frac{x_0^2}{a^2} \) into the ellipse equation:
\[ \frac{x_0^2}{a^2} + \frac{x_0^2}{a^2} = 1 \implies \frac{2x_0^2}{a^2} = 1 \implies x_0^2 = \frac{a^2}{2} \] And from \( \frac{x_0^2}{a^2} = \frac{y_0^2}{b^2} \), we get \( \frac{1}{2} = \frac{y_0^2}{b^2} \implies y_0^2 = \frac{b^2}{2} \).

Step 4: Use the hyperbola equation. Since \( (x_0, y_0) \) also lies on the hyperbola: \[ x_0^2 - y_0^2 = c^2 \]
Substitute the values we found for \( x_0^2 \) and \( y_0^2 \): \[ \frac{a^2}{2} - \frac{b^2}{2} = c^2 \] \[ a^2 - b^2 = 2c^2 \]