Question

1. write the equation and sketch the parabola with a focus at (0,1) and a directrix line at y = 9, one point at (8,1) and additional point at (-4,4) (20 points).

Explanation:

Step1: Recall the definition of a parabola

A parabola is the set of all points \((x,y)\) such that the distance from \((x,y)\) to the focus \((0,1)\) is equal to the distance from \((x,y)\) to the directrix \(y = 9\).
The distance formula is \(d=\sqrt{(x - x_1)^2+(y - y_1)^2}\). The distance from \((x,y)\) to \((0,1)\) is \(\sqrt{(x - 0)^2+(y - 1)^2}\), and the distance from \((x,y)\) to the line \(y = 9\) is \(|y - 9|\).
So, \(\sqrt{x^{2}+(y - 1)^{2}}=|y - 9|\).

Step2: Square both sides of the equation

Squaring both sides, we get \(x^{2}+(y - 1)^{2}=(y - 9)^{2}\).
Expand the squares: \(x^{2}+y^{2}-2y + 1=y^{2}-18y + 81\).

Step3: Simplify the equation

Cancel out \(y^{2}\) terms: \(x^{2}-2y + 1=-18y + 81\).
Move the \(y\) - terms to one side: \(x^{2}=-18y + 81+2y - 1\).
Combine like terms: \(x^{2}=-16y+80\).
Solve for \(y\): \(y =-\frac{1}{16}x^{2}+5\).

Answer:

The equation of the parabola is \(y =-\frac{1}{16}x^{2}+5\). To sketch the parabola:

• The vertex of the parabola \(y = ax^{2}+b\) (in our case \(a =-\frac{1}{16}\), \(b = 5\)) is at \((0,5)\). Since \(a<0\), the parabola opens downwards.
• The focus is at \((0,1)\) and the directrix is \(y = 9\).
• Plot the vertex \((0,5)\), focus \((0,1)\), directrix \(y = 9\), and the given points \((8,1)\) and \((- 4,4)\). For \(x = 8\), \(y=-\frac{1}{16}\times64 + 5=-4 + 5 = 1\); for \(x=-4\), \(y=-\frac{1}{16}\times16 + 5=-1 + 5 = 4\). Then draw a smooth curve passing through these points with the vertex at \((0,5)\) and opening downwards.