Question

If \(S = \{(x,y)\;|\; x = 2\cos t + 3\sin t,\; y = 3\cos t + 2\sin t,\; t \in \mathbb{R}\}\), then the points of \(S\) lie on the curve: 

• A. \(13x^{2}+24xy+13y^{2}=25\)
• B. \(x^{2}+y^{2}+12x+12y+4=0\)
• C. \(13x^{2}-24xy+13y^{2}=25\)
• D. \(x^{2}-2xy+y^{2}=8\)

Hint:

Eliminate parameters using matrix inversion when x,y are linear in sin and cos.

Answer 1

The Correct Option is C

Concept: We eliminate parameter \(t\) using trigonometric identities and express in quadratic form.

Step 1: Write system in matrix form.
\[ x=2\cos t+3\sin t,\quad y=3\cos t+2\sin t \] \[ \begin{bmatrix} x y \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} \cos t \\ \sin t \end{bmatrix} \]

Step 2: Invert transformation.
Determinant: \[ |A|=4-9=-5 \] \[ \cos t=\frac{2x-3y}{-5}, \quad \sin t=\frac{-3x+2y}{-5} \]

Step 3: Use identity \(\sin^2 t+\cos^2 t=1\).
\[ \frac{(2x-3y)^2+(3x-2y)^2}{25}=1 \] Expand: \[ (4x^2-12xy+9y^2)+(9x^2-12xy+4y^2)=25 \] \[ 13x^2-24xy+13y^2=25 \]