Q82 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics

The equation of the locus of a point $(2 \cos \theta-3,3 \sin \theta-4)$ is

  • A
    $9 x^2+4 y^2+54 x+32 y+181=0$
  • B
    $4 x^2+9 y^2+54 x+32 y+109=0$
  • C
    $9 x^2+4 y^2-54 x+32 y+109=0$
  • Correct answer
    D
    $9 x^2+4 y^2+54 x+32 y+109=0$

Solution

Given point $(2 \cos \theta-3,3 \sin \theta-4)=(x, y)$ Let $$ \begin{aligned} & \Rightarrow \quad \because \cos \theta=\frac{x+3}{2} \text { and } \sin \theta=\frac{y+4}{3} \\ & \Rightarrow \quad \cos ^2 \theta+\sin ^2 \theta=1 \\ & \Rightarrow \quad\left(\frac{x+3}{2}\right)^2+\left(\frac{y+4}{3}\right)^2=1 \\ & \Rightarrow \quad 9\left(x^2+6 x+9\right)+4\left(y^2+8 y+16\right)=36 \\ & \Rightarrow \quad 9 x^2+4 y^2+54 x+32 y+145=36 \\ & \Rightarrow \quad 9 x^2+4 y^2+54 x+32 y+109=0 \end{aligned} $$

Hence, the correct answer is option: 'D'.

More Ellipse Questions