Problem 39 - College Algebra

Question

Find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$(y+3)^{2}=12(x+1)$$

Step-by-Step Solution

Answer

The vertex is \((-1, -3)\), the focus of the parabola is \((2, -3)\) and the equation of the directrix of the parabola is \(x = -4\). The parabola opens to the right.

Step 1: Determining the vertex

Compare the given equation \((y+3)^2=12(x+1)\) with standard form \((y-k)^2=4p(x-h)\). From the comparison, it can be inferred that the vertex \((h, k)\) of the parabola is \((-1, -3)\).

Step 2: Calculating the value of p

The coefficient of \(x\) in the given equation is \(12 = 4p\). Solving this equation for \(p\) gives \(p = 3\). Note that since p > 0, the parabola opens to the right.

Step 3: Finding the focus of the parabola

By definition, the focus \((h',k')\) of the parabola is \((h+p, k) = (-1+3, -3) = (2, -3)\).

Step 4: Calculating the equation of directrix

Equation of directrix is \(x=h-p = -1-3 = -4\).

Step 5: Sketching the parabola

Plot the vertex, focus and the directrix of the parabola. Draw a smooth curve passing through the vertex that bends and opens towards the focus while keeping away from the directrix.