The cartesian equation of the curve $x=3+5 \cos \theta, y=2+5 \sin \theta$ is $(0 \leq \theta \leq 2 \pi)$

The cartesian equation of the curve $x=3+5 \cos \theta, y=2+5 \sin \theta$ is $(0 \leq \theta \leq 2 \pi)$
  1. $x^{2}+y^{2}-6 x+4 y-12=0$
  2. $x^{2}+y^{2}+6 x+4 y+12=0$
  3. $x^{2}+y^{2}+6 x-4 y+12=0$
  4. $x^{2}+y^{2}-6 x-4 y-12=0$

Solution

We have $\frac{x-3}{5}=\cos \theta$ and $\frac{y-2}{5}=\sin \theta$ $\therefore \cos ^{2} \theta+\sin ^{2} \theta=1$ gives $\begin{aligned} &\left(\frac{x-3}{5}\right)^{2}+\left(\frac{y-2}{5}\right)^{2}=1 \\ \therefore & x^{2}-6 x+9+y^{2}-4 y+4=25 \\ \therefore & x^{2}+y^{2}-6 x-4 y-12=0 \end{aligned}$

Asked in: MHT CET 2020 (20 Oct Shift 2)

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