91Ó°ÊÓ

The given parabola has vertex $\left(0,1\right)$ and focus$\left(0,5\right)$.

For two different forms of equations of parabola stated below, use the following key-concept to find vertex, axis of symmetry, focus, directrix, direction of opening of parabola and length of latus rectum.

$\boxed{\begin{array}{ccc}\text{Formofequations}& y=a{\left(x−h\right)}^2+k& x=a{\left(y−k\right)}^2+h\\ \text{Vertex}& \left(h,k\right)& \left(h,k\right)\\ \text{Axisofsymmetry}& x=h& y=k\\ \text{Focus}& \left(h,k+\frac{1}{4a}\right)& \left(h+\frac{1}{4a},k\right)\\ \text{Directrix}& y=k−\frac{1}{4a}& x=h−\frac{1}{4a}\\ \text{Directionofopening}& \begin{array}{l}\text{upwardif}a>0,\\ \text{downwardif}a<0\end{array}& \begin{array}{l}\text{rightif}a>0,\\ \text{leftif}a<0\end{array}\\ \text{Lengthoflatusrectum}& \left|\frac{1}{a}\right|\text{units}& \left|\frac{1}{a}\right|\text{units}\end{array}}$

From the given vertex $\left(0,1\right)$and focus $\left(0,5\right)$ it can be interpreted that the x-coordinate remains unchanged and y-coordinate changes. Therefore, vertex of the parabola will be

$\left(h,k\right)=\left(0,1\right)$.

As, y-coordinate changes therefore the equation of the parabola in standard form will be like:

$y=a{\left(x-h\right)}^2+k$

Plugging for vertex $\left(h,k\right)=\left(0,1\right)$in $y=a{\left(x-h\right)}^2+k$.

$\begin{array}{l}y=a{\left(x−\left(0\right)\right)}^2+\left(1\right)\\ y=a{x}^2+\mathrm{1....}\left(1\right)\end{array}$

Plugging for vertex $\left(h,k\right)=\left(0,1\right)$in focus $\left(h,k+\frac{1}{4a}\right)$ are equating.

$\begin{array}{c}\left(h,k+\frac{1}{4a}\right)=\left(0,5\right)\mathrm{....}\left(\text{Given}\right)\\ \left(0,1+\frac{1}{4a}\right)=\left(0,5\right)\\ 1+\frac{1}{4a}=5\\ \frac{1}{4a}=4\\ a=\frac{1}{16}\end{array}$

Plugging $a=\frac{1}{16}$ in (1) gives $y=\frac{1}{16}{x}^2+1$ which is the required equation of the parabola.

The graph of the parabola $y=\frac{1}{16}{x}^2+1$ is shown below.

The required equation of the parabola is $y=\frac{1}{16}{x}^2+1$.