The parametric equations of the curve $x^2+y^2+a x+b y=0$ are
The parametric equations of the curve $x^2+y^2+a x+b y=0$ are
- $x=\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta$
- $x=\frac{\mathrm{a}}{2}-\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=\frac{\mathrm{b}}{2}-\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta$
- $x=-\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=-\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta$
- $x=-\frac{\mathrm{a}}{2}-\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=-\frac{\mathrm{b}}{2}-\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta$
Solution
$\begin{aligned}
& x^2+y^2+\mathrm{a} x+\mathrm{b} y=0 \\
& \Rightarrow\left(x+\frac{\mathrm{a}}{2}\right)^2+\left(y+\frac{\mathrm{b}}{2}\right)^2=\frac{\mathrm{a}^2+\mathrm{b}^2}{4}
\end{aligned}$
Comparing with $(x-\mathrm{h})^2+(y-\mathrm{k})^2=\mathrm{r}^2$, we get
$\mathrm{h}=-\frac{\mathrm{a}}{2}, \mathrm{k}=-\frac{\mathrm{b}}{2}, \mathrm{r}=\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}}$
$\therefore \quad$ The parametric equations are
$x=-\frac{\mathrm{a}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \cos \theta, y=-\frac{\mathrm{b}}{2}+\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{4}} \sin \theta$
Asked in: MHT CET 2023 (13 May Shift 1)
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