Q33. - Algebra 2

Question

Use the equation $x=3y^2+4y+1$ to find the axis of symmetry.

Step-by-Step Solution

Answer

The axis of symmetry is $y=-\frac{2}{3}$.

Step 1. Write down the given information.

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

Step 2. Concept used.

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.

 $\boxed{\begin{array}{ccc}\text{Form of equations}& y=a{\left(x-h\right)}^2+k& x=a{\left(y-k\right)}^2+h\\ \text{Vertex}& \left(h,k\right)& \left(h,k\right)\\ \text{Axis of symmetry}& x=h& y=k\\ \text{Focus}& \left(h,k+\frac{1}{4a}\right)& \left(h+\frac{1}{4a},k\right)\\ \text{Directrix}& y=k-\frac{1}{4a}& x=h-\frac{1}{4a}\\ \text{Direction of opening}& \begin{array}{l}\text{upward if }a>0,\\ \text{downward if }a<0\end{array}& \begin{array}{l}\text{right if }a>0,\\ \text{left if }a<0\end{array}\\ \text{Length of latus rectum}& \left|\frac{1}{a}\right|\text{units}& \left|\frac{1}{a}\right|\text{units}\end{array}}$

Step 3. Convert the given equation to standard form.

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

 $\begin{array}{c}x=3y^2+4y+\mathrm{1....}\left(\text{Given}\right)\\ x=3\left(y^2+\frac{4}{3}y+\frac{1}{3}\right)\\ x=3\left(y^2+\frac{4}{3}y+\frac{1}{3}+{\left(\frac{2}{3}\right)}^2-{\left(\frac{2}{3}\right)}^2\right)\mathrm{....}\left(\begin{array}{l}\text{ Add and }\\ \text{subtract }{\left(\frac{2}{3}\right)}^2\end{array}\right)\\ x=3\left(y^2+\frac{4}{3}y+{\left(\frac{2}{3}\right)}^2\right)+3\left(\frac{1}{3}-{\left(\frac{2}{3}\right)}^2\right)\\ x=3{\left(y+\frac{2}{3}\right)}^2-\frac{1}{3}\mathrm{....}\left(\text{Standard form}\right)\end{array}$

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

Step 4. Evaluating equations of axis of symmetry.

The equation of axis of symmetry for the given equation using the concept stated above is written as:

 $\begin{array}{l}y=k\\ y=-\frac{2}{3}\mathrm{....}\left(\because k=-\frac{2}{3}\right)\end{array}$

Hence, the axis of symmetry is $y=-\frac{2}{3}$.

Step 5. Conclusion.

The axis of symmetry is $y=-\frac{2}{3}$.