Mathematics - Parabola Question with Solution | TestHub

$\mathrm{y}={\mathrm{x}}^2+\mathrm{px}-3$ Let $\mathrm{P}(\alpha ,0),\mathrm{Q}(\beta ,0),\mathrm{R}(0,-3)$ Circle with centre $(-1,-1)$ is $(x+1{)}^2+(y+1{)}^2=r^2$ Passes through $(0,-3)$ $\begin{array}{rl}& 1^2+(-2{)}^2=r^2]\\ & {\mathrm{r}}^2=5\end{array}$ $(x+1{)}^2+(y+1{)}^2=5$ Put $\mathrm{y}=0$ $\begin{array}{rl}& (x+1{)}^2=5-1\\ & (x+1{)}^2=4\\ & x+1=\pm 2\\ & x=1\text{ or }x=-3\\ & \therefore P(1,0)\text{ and }Q(-3,0)\end{array}$ Area of $△\mathrm{PQR}=\frac{1}{2}\left|\begin{array}{ccc}1& 0& 1\\ -3& 0& 1\\ 0& -3& 1\end{array}\right|=6$