Solution:

No. of points are k = 3,

Degree of polynomial $(\mathrm{n})=\mathrm{k}-1=2$

$\therefore$ Quadratic Curve

$P(t)=P_{0}(1-t)^{2}+2 P_{1}(1-t) t+P_{2} t^{2}$

$P_{x}=4(1-t)^{2}+2(0)(1-t) t+2 t^{2}$

$P_{x}=4(1-t)^{2}+2 t^{2}$

$\therefore P_{x}=4-8 t+6 t^{2}$

$P_{y}=2(1-t)^{2}+2(0)(1-t) t+8 t^{2}$

$P_{y}=2(1-t)^{2}+8 t^{2}$

$P_{y}=2\left[1-2 t+t^{2}\right]+8 t^{2}=2-4 t+2 t^{2}+8 t^{2}$

$\therefore P_{y}=2-4 t+10 t^{2}$