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Chapter 6: Problem 42

Find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $$y^{2}-4 y-4 x=0$$

Short Answer

Expert verified
The vertex of the parabola is (0, 2), the focus is at (0, 2.25), and the directrix is the line \( y = 1.75 \). The parabola opens downward, with its vertex at (0, 2), passing through the focus, and the directrix is the horizontal line at \( y = 1.75 \).

Step by step solution

01

Rewrite to Standard Form

To write the given equation \(y^{2}-4 y-4 x=0\) to the standard form, complete the square for the y terms. Add and subtract \( (4/2)^2 = 4 \) inside the equation to obtain: \(y^{2}-4 y+4-4-4x=0\), which simplifies to \(-(y-2)^2-4x=0\) or \((y-2)^2=-4x\), or \((y-2)^2=4(-x)\). Therefore, the standard form of the parabola equation is \((y-2)^2=4(-(x-h))\). Here a=-1, h=0 and k = 2.
02

Identify Vertex, Focus and Directrix

The vertex of the parabola is at point (h, k) = (0, 2). The focus of the parabola, represented by (h, k - 1/4a), is at point (0, 2-1/4(-1)) = (0, 2.25). The directrix is represented by the equation y = k + 1/4a, which simplifies to y = 2 + 1/4*(-1) = 1.75. Thus, the focus is (0, 2.25) and directrix is \( y = 1.75 \).
03

Sketch the Parabola

Plot the vertex at point (0,2). Sketch a vertical line at y = 1.75 to represent the directrix. Mark the focus point at (0, 2.25). Draw the parabola opening downwards with its vertex at (0,2), passing through the focus, and rest on the directrix.

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