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Chapter 7: Problem 43

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. $$x^{2}-2 x-4 y+9=0$$

Short Answer

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

Step by step solution

01

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 \((x-1)^2 = 4(y-2)\).
02

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.
03

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.

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Most popular questions from this chapter

Find the standard form of the equation of each parabola satisfying the given conditions. Focus: \((0,15) ;\) Directrix: \(y=-15\)

Graph each relation. Use the relation's graph to determine its domain and range. \(\frac{y^{2}}{16}-\frac{x^{2}}{9}=1\)

Find the standard form of the equation of each parabola satisfying the given conditions. Focus: \((0,20) ;\) Directrix: \(y=-20\)

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? $$y^{2}-2 y-x-5=0$$

Isolate the terms involving \(y\) on the left side of the equation: $$ y^{2}+2 y+12 x-23-0 $$ Then write the equation in an equivalent form by completing the square on the left side.
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