Question

Let $P = \left(\frac{15}{2}(\csc \theta + \sin \theta), \; 8(\csc \theta - \sin \theta)\right)$, where $\theta$ is a variable parameter. Then the locus of $P$ is

• A. $\frac{x^2}{15} - \frac{y^2}{16} = 1$
• B. $\frac{x^2}{256} - \frac{y^2}{225} = 1$
• C. $\frac{x^2}{225} + \frac{y^2}{256} = 1$
• D. $\frac{x^2}{225} - \frac{y^2}{256} = 1$
• E. $\frac{x^2}{16} + \frac{y^2}{30} = 1$

Hint:

Convert parametric trig forms into algebraic form using identities like $a^2-b^2$.

Answer 1

The Correct Option is D

Concept:
• Use identity: $\csc^2\theta - \sin^2\theta = \frac{1}{\sin^2\theta} - \sin^2\theta$
• Convert parametric form into Cartesian equation

Step 1: Let coordinates be
\[ x = \frac{15}{2}(\csc\theta + \sin\theta), \quad y = 8(\csc\theta - \sin\theta) \]

Step 2: Form expressions
\[ \frac{2x}{15} = \csc\theta + \sin\theta, \quad \frac{y}{8} = \csc\theta - \sin\theta \]

Step 3: Add and subtract
Adding: \[ \frac{2x}{15} + \frac{y}{8} = 2\csc\theta \] Subtracting: \[ \frac{2x}{15} - \frac{y}{8} = 2\sin\theta \]

Step 4: Use identity
\[ (\csc\theta)^2 - (\sin\theta)^2 = 1 \] \[ \left(\frac{1}{2}\left(\frac{2x}{15} + \frac{y}{8}\right)\right)^2 - \left(\frac{1}{2}\left(\frac{2x}{15} - \frac{y}{8}\right)\right)^2 = 1 \]

Step 5: Simplify
\[ \frac{x^2}{225} - \frac{y^2}{256} = 1 \] Final Conclusion:
Option (D)