How do you find an equation of the parabola with a vertex of (-1,2) and focus of (-1,0)?

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Answer 1 · 2015-02-13T18:53:01

The answer is: #y=-1/8x^2-1/4x+15/8#

First of all some formulas:

#y=ax^2+bx+c# is the equation of a parabola with axis of symmetry parallel to the y-axis;

#V(-b/(2a),-Delta/(4a))# where #Delta=b^2-4ac#, is the vertex;

#F(-b/(2a),(1-Delta)/(4a))# is the focus;

#x=-b/(2a)# is the axis of symmetry;

#y=(-1-Delta)/(4a)# is the directrix.

So we have to solve this system of equations:

#-b/(2a)=-1#

#-Delta/(4a)=2#

#(1-Delta)/(4a)=0#

Than:

#b=2a#

#Delta=-8a#

#Delta=1#

And:

#a=-1/8#

#b=-1/4#

#Delta=1#

From the last one we have to find the value of #c#

#b^2-4ac=1rArr1/16-4*(-1/8)*c=1rArr1/16+1/2c=1rArr#

#1/2c=1-1/16rArrc=2*15/16rArrc=15/8#.

The equation becomes: #y=-1/8x^2-1/4x+15/8#