How do you find the vertex, focus and directrix of #y^2+4y+8x-12=0#?

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Answer 1 · 2016-07-20T08:25:20

Vertex#=(2,-2)#, Focus#=(0,-2)#

Eqn. of Dir. #x-4=0#

Given eqn. #y^2+4y=-8x+12#

Completing square, #y^2+4y+4=4-8x+12#

#:. (y+2)^2=-8x+16=4*(-2)(x-2)#

Let, #(y+2)=Y, and, (x-2)=X#, so the eqn. becomes,

#Y^2=4aX............(1)#, which is the std. eqn. of a Parabola, with

#a=-2#.

Now, we know that the parabola #(1)#, the Vertex is (0,0), Focus is #S(a,0)# and eqn. of directrix, is #X+a=0#.

Vertex :-

#(X,Y)=(0,0)rArr(x-2,y+2)=(0,0)rArr(x,y)=(2,-2)#.

Focus :-

#S(X,Y)=S(a,0)=S(-2,0)rArrS(x-2,y+2)=S(-2,0)rArrS(x,y)=S(0,-2)#.

Directrix :-

Eqn. is,#X+a=0rArrX-2=0rArrx-2-2=0rArrx-4=0#.

Hope, this is helpful! Enjoy Maths.!