What are the vertex, focus and directrix of # y=4x^2 + 5x + 7 #?

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Answer 1 · 2018-07-03T04:28:18

Given equation:

#y=4x^2+5x+7#

#y=4(x^2+5/4x)+7#

#y=4(x^2+5/4x+25/64)-25/64+7#

#y=4(x+5/8)^2+423/64#

#(x+5/8)^2=1/4(y-423/64)#

Comparing above equation with the standard form of parabola #X^2=4aY# we get

#X=x+5/8, \ Y=y-423/64, a=1/16#

Vertex of Parabola

#X=0, Y=0#

#x+5/8=0, y-423/64=0#

#x=-5/8, y=423/64#

#(-5/8, 423/64)#

Focus of parabola

#X=0, Y=a#

#x+5/8=0, y-423/64=1/16#

#x=-5/8, y=427/64#

#(-5/8, 427/64)#

Directrix of parabola

#Y=-a#

#y-423/64=-1/16#

#y=419/64#