Problem 46 - 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. $$y^{2}-2 y-8 x+1=0$$

Step-by-Step Solution

Answer

The equation in standard form is \((y-1)^{2} = 8x\). The vertex of the parabola is at (0,1). The focus is at (2,1) and the directrix is \(x = -2\).

Step 1: Convert to standard form

To convert the equation to standard form, we complete the square on \(y\). We will rewrite the equation by grouping \(y\) terms together: \(y^{2}-2y=8x-1\). To complete the square for y, we add \((-2/2)^2 = 1\) on both sides to get \(y^{2}-2y+1=8x\). So, in standard form, the equation is \((y-1)^{2}=8x\).

Step 2: Find the vertex

In an equation of from \((y-k)^{2}=4a(x-h)\), the vertex is given by \((h,k)\). Comparing this to our equation, we see that \(h=0\) and \(k=1\), so the vertex is \((0,1)\).

Step 3: Find the focus

The focus is at location \((h+a, k)\). In our case, a is determined by the equation \(4a=8 => a=2\). So, the focus is at \((0+2,1) = (2,1)\).

Step 4: Find the directrix

The directrix is at location \(x=h-a\). In our case, this gives \(x=-2\).

Step 5: Graph the Parabola

You start by plotting the vertex, focus and directrix on a graph. Next draw a smooth curve to form the parabola. It should pass through the vertex and focus, and be mirrored across the directrix.