Question

If the curves $$ 2x^2 + ky^2 = 30 \quad \text{and} \quad 3y^2 = 28x $$ cut each other orthogonally, then \( k = \) 

• A. \( 5 \)
• B. \( 3 \)
• C. \( 2 \)
• D. \( 1 \)

Hint:

For curves intersecting orthogonally, use the condition \( m_1 m_2 = -1 \), where \( m_1 \) and \( m_2 \) are the slopes of the tangents at the intersection point.

Answer 1

The Correct Option is D

Step 1: Find the Slopes of the Tangents
Differentiate the first equation: \[ 2x^2 + ky^2 = 30. \] \[ 4x + 2ky \frac{dy}{dx} = 0. \] \[ \frac{dy}{dx} = -\frac{2x}{ky}. \] Differentiate the second equation: \[ 3y^2 = 28x. \] \[ 6y \frac{dy}{dx} = 28. \] \[ \frac{dy}{dx} = \frac{14}{3y}. \] Step 2: Use the Orthogonality Condition
For curves to intersect orthogonally: \[ m_1 m_2 = -1. \] Substituting values: \[ \left(-\frac{2x}{ky} \right) \times \left( \frac{14}{3y} \right) = -1. \] \[ \frac{-28x}{3ky^2} = -1. \] \[ 28x = 3ky^2. \] Step 3: Solve for \( k \)
Using \( 3y^2 = 28x \): \[ k(28x) = 28x. \] \[ k = 1. \] Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{1}. \]