Respuesta :

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]

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