Find the equation of a parabola (in general form)

Find the equation of the parabola with axis parallel to the $y$-axis, passing through $(1/2,-5/2),(3/2,-9/4)$ and $(-7/2,3/2)$.

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possible duplicate of Find the equation of the parabola – Zia Aug 31 '13 at 10:13
This is not a duplicate . the idea is different although this also deals with finding equation of parabola – Harish Kayarohanam Aug 31 '13 at 10:16
What is the "axis" of a parabola? Is this a common term? – Stefan Smith Aug 31 '13 at 12:52

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} {\rm y}\pars{x}&= A\pars{x - \half}\pars{x - {3 \over 2}} +B\pars{x - {3 \over 2}}\pars{x + {7 \over 2}} +C\pars{x - \half}\pars{x + {7 \over 2}} \end{align}

In order to determine $\ds{A, B , C}$ use $${\rm y}\pars{-\,{7 \over 2}} = {3 \over 2}\,,\qquad {\rm y}\pars{\half} = -\,{5 \over 2}\,,\qquad {\rm y}\pars{3 \over 2} = -\,{9 \over 4}$$

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The above answer is a good shortcut, but by convention it's as follows:

With axis parallel to the y axis, if the vertex of the parabola is on the origin, then the equation is $x^2=4ay$. But when you shift the parabola on a vertex with coordinates $(h,k)$, the equation becomes $(x-h)^2=4a(y-k)$. Substitute the values you gave to find h,k and a (three equations, three unknowns).

But I would also recommend going by $y=ax^2+bx+c$, since the convention might get messy.

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Using the standard form for the equation of the parabola doesn't create as much trouble as you might think, since the $\ 4ak \$ term will cancel between any pair of equations, reducing them to having two unknowns, $\ a \$ and $\ h \$ . You can then go back and pick up $\ k \$ . – RecklessReckoner Mar 28 '14 at 3:42

HINT:(Idea behind the problem) A parabola is a graph of a quadratic function

$$\displaystyle\boxed{y =ax^2 + bx + c}$$

substitute 3 points given , you will get 3 equations in a,b,c and from there find a , b , c solving the 3 equations .and then substitute this a,b,c in the equation above and that will be your answer .

refer an example : refer in case you have doubts and want example

solution from wolfram alpha : wolfram alpha

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Your parabola has axis parallel to the $x$-axis. – walcher Aug 31 '13 at 9:46
No . this is only for axis parallel to y axis . see en.wikipedia.org/wiki/File:Quadratic_function.png – Harish Kayarohanam Aug 31 '13 at 9:50
oh, is it different from x^2 +dx + ey + f = 0? – mona Aug 31 '13 at 9:51
My bad, I thought by axis you meant directrix. – walcher Aug 31 '13 at 9:56
@mona I didn't get what you are asking in comment 3. – Harish Kayarohanam Aug 31 '13 at 10:41

Standard form: $(x+20)^2 + (y+24)^2 = (\dfrac{185}{2})^2$

General form: $2x^2 + 2y^2 - 20x -24y - 63$

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