Question

1. write the equation and sketch the parabola who is facing upward and has point located at (-2,9), a y - intercept at (0,6), a vertex at (2,5), and a directrix line that is 1 unit away from the vertex (20 points)

Explanation:

Step1: Recall the standard form of a parabola

The standard form of a parabola that opens upward is \((x - h)^2=4p(y - k)\), where \((h,k)\) is the vertex. Given the vertex \((h,k)=(2,5)\), the equation becomes \((x - 2)^2 = 4p(y - 5)\).

Step2: Determine the value of \(p\)

Since the parabola is facing upward and the directrix is 1 unit away from the vertex. For a parabola opening upward, the distance from the vertex to the directrix is \(p\). So \(p = 1\).

Step3: Substitute \(p\) into the equation

Substitute \(p = 1\) into \((x - 2)^2=4p(y - 5)\). We get \((x - 2)^2=4(y - 5)\).

Step4: Sketching the parabola (brief description)

• Vertex: Plot the point \((2,5)\).
• Directrix: The directrix is \(y=5 - 1=4\), draw the horizontal line \(y = 4\).
• Using the equation:
• Let \(x=2\), then \(y = 5\) (vertex).
• Let \(x=0\): \((0 - 2)^2=4(y - 5)\), \(4 = 4(y - 5)\), \(y-5 = 1\), \(y=6\).
• Let \(x = 4\): \((4 - 2)^2=4(y - 5)\), \(4=4(y - 5)\), \(y = 6\).
• The \(y\)-intercept: Let \(x = 0\), from \((x - 2)^2=4(y - 5)\), \(4=4(y - 5)\), \(y=6\) (not \((0,6)\) is a point on the parabola, but we can also check the given \(y\)-intercept condition. Wait, we can also use the general property. Since \(p = 1\), focus is at \((2,5 + 1)=(2,6)\).
• Plot several points and draw a smooth curve passing through them, opening upward with vertex at \((2,5)\) and directrix \(y = 4\).

Answer:

The equation of the parabola is \((x - 2)^2=4(y - 5)\)