How do you find the vertex, focus and directrix of #16(y+3)^2 = (x - 2)#?

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Answer 1 · 2017-07-29T12:30:12

Vertex is at # (2, -3) #, focus is at #(129/64, -3) # and
directrix is #x = 127/64#

# 16(y+3)^2= (x-2) or (y+3)^2 = 1/16 (x-2)# Comparing with

standard equation of parabola opening rightward

#(y-k)^2 = 4a (x-h) ; h= 2 , k= -3 , a = 1/64 # .

The vertex is at # (h,k) or (2, -3) # , focus and directris is at

equidistance from vertex situated at oppste sides,

Focus is at #((h+a),k or( (2+1/64), -3) or (129/64, -3) #

Directrix is #x =(h-a) or x= 2-1/64 or x = 127/64#

graph{16(y+3)^2 =(x-2) [-10, 10, -5, 5]} [Ans]