Question

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

Answer 1

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`