What is the equation in standard form of the parabola with a focus at (-10,-47) and a directrix of y= -5?

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Answer 1 · 2016-10-05T09:32:54

#(x+10)^2=-84(y+26)#, revealing that

the vertex is V(-10, -26) and the axis is #x=-10 (downwards)..

Use

distance PS of P(x, y) on the parabola from its focus #S(-10, -47)#

= its perpendicular distance PN from the directrix #y=-5#

#PS=sqrt((x+10)^2+(y+47)^2) = PN =+-(y-(-5+)#.

Squaring and equating,

#(x+10)^2=-84y-2286# In the standard form,

#(x+10)^2=-84(y+26)# revealing that

the vertex is V(-10, -26) and the axis is #x=-10 (downwards).