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: $y = - \frac{1}{8} x^{2} - \frac{1}{4} x + \frac{15}{8}$ $y=-1/8x^2-1/4x+15/8$

First of all some formulas:

$y = a x^{2} + b x + c$ $y=ax^2+bx+c$ 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: $y=-1/8x^2-1/4x+15/8$