Question

graph the parabola given by the following equation using the focus and the directrix.
$(y - 1)^2 = -20(x + 1)$

Explanation:

Step1: Recall the standard form of a parabola

The standard form of a parabola that opens horizontally is \((y - k)^2 = 4p(x - h)\), where \((h,k)\) is the vertex, \(p\) is the distance from the vertex to the focus (and also from the vertex to the directrix). If \(p>0\), it opens to the right; if \(p < 0\), it opens to the left.

Step2: Identify \(h\), \(k\), and \(p\) from the given equation

Given the equation \((y - 1)^2=-20(x + 1)\), we can rewrite it as \((y - 1)^2=4p(x - (- 1))\). Comparing with the standard form \((y - k)^2 = 4p(x - h)\), we have:

• \(h=-1\), \(k = 1\) (so the vertex is \((h,k)=(-1,1)\))
• \(4p=-20\), so we solve for \(p\): \(p=\frac{-20}{4}=-5\)

Step3: Find the focus

The focus of a horizontally opening parabola is at \((h + p,k)\). Substituting \(h=-1\), \(p = - 5\), and \(k = 1\):
Focus \(=( - 1-5,1)=(-6,1)\)

Step4: Find the directrix

The directrix of a horizontally opening parabola is the vertical line \(x=h - p\). Substituting \(h=-1\) and \(p=-5\):
Directrix: \(x=-1-(-5)=-1 + 5 = 4\), so the directrix is the line \(x = 4\)

Step5: Graphing the parabola

1. Plot the vertex \((-1,1)\).
2. Plot the focus \((-6,1)\) (since \(p=-5<0\), the parabola opens to the left, towards the focus direction).
3. Draw the directrix \(x = 4\) (a vertical line).
4. To sketch the parabola, we know that any point on the parabola is equidistant from the focus and the directrix. We can use the vertex and the direction (opening to the left) to draw the parabola. For example, we can find a few more points. Let's take \(y=1 + 10\) (i.e., \(y = 11\)) and \(y=1-10\) (i.e., \(y=-9\)) and find the corresponding \(x\) values.

Substitute \(y = 11\) into \((y - 1)^2=-20(x + 1)\):
\((11 - 1)^2=-20(x + 1)\)
\(100=-20(x + 1)\)
\(x+1=\frac{100}{-20}=-5\)
\(x=-6\)
So the point \((-6,11)\) is on the parabola.
Substitute \(y=-9\) into \((y - 1)^2=-20(x + 1)\):
\((-9 - 1)^2=-20(x + 1)\)
\(100=-20(x + 1)\)
\(x + 1=-5\)
\(x=-6\)
So the point \((-6,-9)\) is on the parabola. We can use these points (and the symmetry about the horizontal line \(y = 1\)) to draw the parabola opening to the left with vertex at \((-1,1)\), focus at \((-6,1)\), and directrix \(x = 4\).

Answer:

To graph the parabola \((y - 1)^2=-20(x + 1)\):

• Vertex: \((-1,1)\)
• Focus: \((-6,1)\)
• Directrix: \(x = 4\)

Sketch the parabola opening to the left (since \(p=-5<0\)) with the vertex at \((-1,1)\), focus at \((-6,1)\), and directrix \(x = 4\), ensuring points on the parabola are equidistant from the focus and directrix. (For a visual graph, plot the vertex, focus, directrix, and use symmetric points like \((-6,11)\) and \((-6,-9)\) to define the curve.)