Q27. - Algebra 2

Question

Identify the coordinates of the vertex and focus, the equations of the axis of symmetry and directrix, and the direction of opening of the parabola with the given equation $x=-4y^2+6y+2$. Then find the length of latus rectum and graph the parabola.

Step-by-Step Solution

Answer

The vertex is $\left(\frac{17}{4},\frac{3}{4}\right)$. Focus is $\left(\frac{67}{16},\frac{3}{4}\right)$. Axis of symmetry is $y=\frac{3}{4}$. The equation of directrix is $x=\frac{69}{16}$. The direction of opening of parabola is towards left. The length of latus rectum is $\frac{1}{4}\text{ unit}$.

Step 1. Write down the given information.

The given equation is $x=-4y^2+6y+2$.

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 $x=-4y^2+6y+2$ is converted to standard form $x=a{\left(y-k\right)}^2+h$ as:

 $\begin{array}{c}x=-4y^2+6y+\mathrm{2....}\left(\text{Given}\right)\\ x=-4\left(y^2-\frac{3}{2}y-\frac{1}{2}\right)\\ x=-4\left(y^2-\frac{3}{2}y-\frac{1}{2}+{\left(\frac{3}{4}\right)}^2-{\left(\frac{3}{4}\right)}^2\right)\mathrm{....}\left(\begin{array}{l}\text{ Add and}\\ \text{subtract }{\left(\frac{3}{4}\right)}^2\end{array}\right)\\ x=-4\left(y^2-\frac{3}{2}y+{\left(\frac{3}{4}\right)}^2\right)+4\left(\frac{1}{2}+{\left(\frac{3}{4}\right)}^2\right)\\ x=-4{\left(y-\frac{3}{4}\right)}^2+\frac{17}{4}\mathrm{....}\left(\text{Standard form}\right)\end{array}$

Comparing $x=-4{\left(y-\frac{3}{4}\right)}^2+\frac{17}{4}$ with $x=a{\left(y-k\right)}^2+h$, $a=-4,h=\frac{17}{4}\text{ and }k=\frac{3}{4}$.

Step 4. Evaluating vertex, focus, equations of axis of symmetry and directrix and direction of opening of parabola.

The vertex and focus, the equations of the axis of symmetry and directrix, and the direction of opening of the parabola are evaluated as:

The vertex is $\left(\frac{17}{4},\frac{3}{4}\right)$.

Focus is evaluated as:

 $\begin{array}{c}\left(h+\frac{1}{4a},k\right)=\left(\frac{17}{4}+\frac{1}{4\left(-4\right)},\frac{3}{4}\right)\\ =\left(\frac{67}{16},\frac{3}{4}\right)\end{array}$

Axis of symmetry is $y=\frac{3}{4}$.

The equation for directrix is evaluated as:

 $\begin{array}{c}x=h-\frac{1}{4a}\\ x=\frac{17}{4}-\frac{1}{4\left(-4\right)}\\ x=\frac{69}{16}\mathrm{....}\left(\text{Directrix}\right)\end{array}$

The direction of opening of parabola is left because $a<0$.

The length of latus rectum is evaluated as:

 $\begin{array}{c}\frac{1}{\left|a\right|}=\frac{1}{\left|-4\right|}\\ =\frac{1}{4}\text{ unit}\end{array}$

Step 5. Sketch the graph of the parabola.

The graph of the parabola is shown below.

Step 6. Conclusion.

The vertex is $\left(\frac{17}{4},\frac{3}{4}\right)$. Focus is $\left(\frac{67}{16},\frac{3}{4}\right)$. Axis of symmetry is $y=\frac{3}{4}$. The equation of directrix is $x=\frac{69}{16}$. The direction of opening of parabola is towards left. The length of latus rectum is $\frac{1}{4}\text{ unit}$.