Question

Find the equation of the parabola having focus (3, 2) and vertex (-1, 2) is

Answer 1

Hint: Find the distance between the focus and vertex using the distance formula and take the value as P. Now substitute the value of vertex and P in the standard form of the equation of parabola in the horizontal axis.

We know that parabola is a U – shaped plane curve where any point is at an equal distance from a fixed straight line which is known as the directrix.
Here we have been the co – ordinates of focus of a parabola as (3, 2).
We know the general equation of parabola as \[{{y}^{2}}=4ax\], which is along the x –axis. Here is the distance between vertex and the focus. Here y – coordinate in focus and vertex is the same. Thus the parabola would be along the x – axis.
Now we need to find the distance between the vertex and focus. We can find the distance using the distance formula. According to the formula,
Distance = \[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
Here, \[\left( {{x}_{2}},{{y}_{2}} \right)\] = focus = (3, 2).
\[\left( {{x}_{1}},{{y}_{1}} \right)\] = vertex = (-1, 2).
\[\therefore \] Distance between vertex and focus \[=\sqrt{{{\left( 3-\left( -1 \right) \right)}^{2}}+{{\left( 2-2 \right)}^{2}}}\]
                                                                      \[\begin{align}
  & =\sqrt{{{4}^{2}}+0} \\
 & =4 \\
\end{align}\]
The general horizontal parabola, center \[\left( {{x}_{0}},{{y}_{0}} \right)\] focus \[\left( {{x}_{0}}{{y}_{0}}+P \right)\] is given as,
\[{{\left( y-{{y}_{0}} \right)}^{2}}=4P\left( x-{{x}_{0}} \right)\]
Thus we got P = 4, which is the distance between vertex and parabola.
\[{{\left( y-{{y}_{1}} \right)}^{2}}=4P\left( x-{{x}_{1}} \right)\] Put, \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( -1,2 \right)\].
\[\begin{align}
  & {{\left( y-2 \right)}^{2}}=4\times 4\left( x-\left( -1 \right) \right) \\
 & {{\left( y-2 \right)}^{2}}=16\left( x+1 \right) \\
\end{align}\]
Thus we got the required equation of the parabola.
i.e. \[{{\left( y-2 \right)}^{2}}=16\left( x+1 \right)\]

Note: If the parabola has a horizontal axis, the standard form of the equation of the parabola is
\[{{\left( y-{{y}_{0}} \right)}^{2}}=4p\left( x-{{x}_{0}} \right)\]
In the case of parabola has vertical axis, the standard form of the equation of the parabola is
\[4p\left( y-{{y}_{0}} \right)={{\left( x-{{x}_{0}} \right)}^{2}}\]