Problem 43 - College Algebra

Question

Convert each equation to standard form by completing the square on \(x\) or \(y .\) Then find the vertex, focus, and directrix of the parabola. Finally, graph the parabola. $$x^{2}-2 x-4 y+9=0$$

Step-by-Step Solution

Answer

The vertex of the parabola is at (1, 2), the focus at (1,3) and the directrix at y=1.

Step 1: Rewriting & Completing the Square

Start by arranging the equation \(x^{2}-2 x-4 y+9=0\) as a quadratic equation. This can be done by isolating the \(x\) terms and constant term on one side: \(x^{2}-2x=4y-9\). \(x^{2}-2x+1 = 4y -9 + 1\), next complete the square on the left side of the equation by adding and subtracting \(\frac{-2}{2}^{2}\), which results in an equation \(x^{2}-2x+1=4y-8\). This can be then rewritten as \((x-1)^2 = 4(y-2)\).

Step 2: Identify characteristics of the parabola

Now, we can identify the vertex, focus and directrix of the parabola by comparing our equation with the standard form of a parabola, which is \((x-h)^2=4p(y-k)\). Here, the vertex (h, k) is (1, 2) and 4p equals 4, so p equals 1. Thus, the vertex is at (1,2), the focus at (1,3) and the directrix at y=1.

Step 3: Graph the parabola

Plot the vertex, focus and directrix line on graphing paper. After that, plot other characteristic points, symmetric with respect to the vertex. By connecting those points with a smooth curve, you can draw the parabola.