Answer:
Step-by-step explanation:
eq. of directrix is y=4 or y-4=0
perpendicular distance of (x,y) from directrix =distance of (x,y) from focus (-3,2)
[tex]| \frac{y-4}{1}|=\sqrt{(x+3)^2+(y-2)^2} \\squaring~both~sides\\y^2-8y+16=(x+3)^2+(y-2)2\\(x+3)^2=y^2-8y+16-(y-2)^2\\(x+3)^2=y^2-8y+16-(y^2-4y+4)\\(x+3)^2=y^2-8y+16-y^2+4y-4\\(x+3)^2=-4y+12\\(x+3)^2=-4(y-3)[/tex]
