91Ó°ÊÓ

The given equation is $x=3y^2+4y+1$.

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.

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

The given equation $x=3y^2+4y+1$ is converted to standard form $x=a{\left(y-k\right)}^2+h$ as:

$\begin{array}{c}x=3y^2+4y+\mathrm{1....}\left(\text{Given}\right)\\ x=3\left(y^2+\frac{4}{3}y+\frac{1}{3}\right)\\ x=3\left(y^2+\frac{4}{3}y+\frac{1}{3}+{\left(\frac{2}{3}\right)}^2−{\left(\frac{2}{3}\right)}^2\right)\mathrm{....}\left(\begin{array}{l}\text{Addand}\\ \text{subtract}{\left(\frac{2}{3}\right)}^2\end{array}\right)\\ x=3\left(y^2+\frac{4}{3}y+{\left(\frac{2}{3}\right)}^2\right)+3\left(\frac{1}{3}−{\left(\frac{2}{3}\right)}^2\right)\\ x=3{\left(y+\frac{2}{3}\right)}^2−\frac{1}{3}\mathrm{....}\left(\text{Standardform}\right)\end{array}$

On comparing $x=3{\left(y+\frac{2}{3}\right)}^2-\frac{1}{3}$with $x=a{\left(y-k\right)}^2+h$, $a=3,h=-\frac{1}{3}\text{and}k=-\frac{2}{3}$.

Hence, the coordinates of the vertex is $\left(-\frac{1}{3},-\frac{2}{3}\right)$.

The coordinates of the vertex is $\left(-\frac{1}{3},-\frac{2}{3}\right)$.