Question

the equation of parabola with the vertex (2,-1)and focus (2,-3)​

Answer 1

Answer:

Step-by-step explanation:

$A(2,-1)\\\\F(2,-3)\\\\AF=a\\\\a=2\\\\Now\;equation\;of\;directrix\;is\\\\y=1\\\\y-1=0\\\\let\;P\;be\;any\;point\;on\;parabola\;and\;M\;be \;its \;foot\;of\;perpendicular\\\\WKT\\\\FP=PM\\\\\sqrt{(x-2)^2+(y+3)^2}=\frac{y-1}{1}\\\\SOBS\\\\x^2+y^2-4x+6y+13=y^2-2y+1\\\\x^2-4x+12+8y=0\\\\(x-2)^2=-8(y+1)$

Hope it helps