Question

How do I convert the standard equation of a parabola to vertex form?

Answer 1

Hint: Problems like these are very simple to solve once we know the different equations and the formulae of the topics in depth and detail. This given problem is the coordinate geometry of sub-topic parabola. The general or the standardised equation of a parabola is generally a quadratic equation and is represented as, \[y=a{{x}^{2}}+bx+c\] . We now need to transform this equation in such a way such that it becomes similar to the general representation of the parabola in vertex form. Some of the general equations of the parabola in vertex form are as follows.

Equation	Vertex	Foci
\[{{y}^{2}}=4ax\]	\[\left( 0,0 \right)\]	\[\left( a,0 \right)\]
\[{{y}^{2}}=-4ax\]	\[\left( 0,0 \right)\]	\[\left( -a,0 \right)\]
\[{{x}^{2}}=4by\]	\[\left( 0,0 \right)\]	\[\left( 0,b \right)\]
\[{{x}^{2}}=-4by\]	\[\left( 0,0 \right)\]	\[\left( 0,-b \right)\]

Complete step by step solution:
Now we start off with the solution to the problem by transforming the general equation of the parabola to one of the type of general equation in vertex form. We do it by trying to rearrange the terms of the quadratic equation and convert it into a perfect square. We do it as follows,
 \[\begin{align}
  & y=a{{x}^{2}}+bx+c \\
 & \Rightarrow y=a{{x}^{2}}+2.\sqrt{a}.\dfrac{b}{2\sqrt{a}}x+c+\dfrac{{{b}^{2}}}{4a}-\dfrac{{{b}^{2}}}{4a} \\
 & \Rightarrow y={{\left( \sqrt{a}x+\dfrac{b}{2\sqrt{a}} \right)}^{2}}+\left( c-\dfrac{{{b}^{2}}}{4a} \right) \\
 & \Rightarrow y-\left( c-\dfrac{{{b}^{2}}}{4a} \right)=a{{\left( x+\dfrac{b}{2a} \right)}^{2}} \\
\end{align}\]
We know rearrange the terms of this above intermediate parabolic equation to get the desired result,
\[\begin{align}
  & \Rightarrow y-\left( c-\dfrac{{{b}^{2}}}{4a} \right)=a{{\left( x+\dfrac{b}{2a} \right)}^{2}} \\
 & \Rightarrow {{\left( x+\dfrac{b}{2a} \right)}^{2}}=4.\left( \dfrac{1}{4a} \right)\left\{ y-\left( c-\dfrac{{{b}^{2}}}{4a} \right) \right\} \\
\end{align}\]
Now, form this above equation we can very clearly say that the quadratic equation has very well been transformed and it looks similar to that of \[{{x}^{2}}=4by\] which is basically the vertex form for the equation. Thus the vertex of our parabola of quadratic form is,
\[\left( -\dfrac{b}{2a},\left\{ c-\dfrac{{{b}^{2}}}{4a} \right\} \right)\]

Note: For problems like these first of all which is important is to remember the general form of the parabola, both in the quadratic form and the vertex form. The conversion from the quadratic form to the vertex from is pretty easy but requires some effort for transforming it to the square form. The last thing that we have to do is observe which of the general equations in vertex form is similar to that of our resultant formed equation.