Question

What is the equation of the parabola described: Vertex at \[\left( 1,6 \right)\] , and focus at \[\left( 2,6 \right)\] ?

Answer 1

Hint: Problems like these are quite easy in general and simple to solve. To solve these types of problems effectively, we need to understand all the key concepts behind the question properly. This particular problem is of topic coordinate geometry and of subtopic parabola. We first need to know all the different types of general forms of parabola that are possible in this field and then we need to compare this problem with the general form to find the solution to the question. Knowing all the general formulae helps us in solving the problem more easily. The general form of the parabolas 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, starting off with the solution of our given problem by writing that, here we first of all need to find the value of the length of the latus rectum which is denoted by \[4a\] . The distance between the vertex and the foci of a parabola is denoted by ‘a’. We now find the value of ‘a’ by finding the distance between the vertex and the foci,
\[\begin{align}
  & a=\sqrt{{{\left( 1-2 \right)}^{2}}+{{\left( 6-6 \right)}^{2}}} \\
 & \Rightarrow a=\sqrt{{{1}^{2}}} \\
 & \Rightarrow a=1 \\
\end{align}\]
Now the value of \[4a\] will be equal to,
\[4a=4\]
Now, in this problem we see that, the vertex is to the left of foci and they are on the same horizontal line. From this we can easily conclude that, the given parabola will be of the form \[{{y}^{2}}=4ax\] . Here applying the change for the origin we get the equation of the parabola as,
\[{{\left( y-k \right)}^{2}}=4a\left( x-h \right)\]
Here \[\left( h,k \right)\] are the co-ordinates of the vertex.
Therefore the equation of the parabola is,
\[\begin{align}
  & {{\left( y-6 \right)}^{2}}=4\left( x-1 \right) \\
 & \Rightarrow {{y}^{2}}+36-12y=4x-4 \\
 & \Rightarrow {{y}^{2}}-12y-4x+40=0 \\
\end{align}\]

Note: Solving this problem requires some basic as well as advanced knowledge of co-ordinate geometry and parabola. We can also solve this particular problem in another method which is by using the concepts of graph. In graph, we plot the vertex and the foci and then plot the figure accordingly. This method is a lengthier process and requires some extra effort of plotting the graph. We must be very careful while doing shifting the origin of the parabola as it is one of the major source of errors.