Question

Write an equation of the parabola that will be open to the left.

Short answer

The vertex is $\left(3,-7\right)$. Focus is $\left(3,-\frac{111}{16}\right)$. Axis of symmetry is $x=3$. The equation of directrix is $y=-\frac{113}{16}$.

Step-by-step solution

Step 1. Write down the given information.

The given equation is $y=4{\left(x-3\right)}^2-7$.

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.

Step 3. Calculation.

From the concept stated above, the parabola will be of the form $x=a{\left(y-k\right)}^2+h$where $a<0$. Therefore, the equation of the parabola is written as:

$\begin{array}{c}x=a{\left(y-k\right)}^2+h\mathrm{....}\left(\begin{array}{l}\text{where}a<0\text{and}\\ \left(h,k\right)\in R\left(\text{Real}\right)\end{array}\right)\\ x=-3{\left(y-3\right)}^2+\mathrm{4....}\left(\begin{array}{l}\text{where}a=-3\text{and}\\ h=4,k=-3\end{array}\right)\end{array}$

Step 4. Conclusion.

The equation of the parabola which will be open to left is $x=-3{\left(y-3\right)}^2+4$.