Question

write the equation for a parabola with a focus at (7, 2) and a directrix at y = -2. y =

Explanation:

Step1: Find the vertex

The vertex \((h,k)\) is halfway between the focus \((7,2)\) and the directrix \(y = - 2\). The \(x\)-coordinate of the vertex is the same as the \(x\)-coordinate of the focus, so \(h=7\). The \(y\)-coordinate of the vertex is the average of the \(y\)-coordinate of the focus and the directrix value: \(k=\frac{2+( - 2)}{2}=\frac{0}{2} = 0\). So the vertex is \((7,0)\).

Step2: Determine the value of \(p\)

The distance between the vertex and the focus (or the vertex and the directrix) is \(|p|\). The focus is at \((7,2)\) and the vertex is at \((7,0)\), so \(p=2 - 0=2\) (since the parabola opens up because the focus is above the directrix).

Step3: Use the standard form of a parabola

The standard form of a parabola that opens up or down is \((x - h)^2=4p(y - k)\). Substituting \(h = 7\), \(k = 0\) and \(p=2\) into the formula: \((x - 7)^2=4\times2\times(y - 0)\), which simplifies to \((x - 7)^2 = 8y\).

Step4: Solve for \(y\)

To get \(y\) by itself, divide both sides of the equation \((x - 7)^2=8y\) by 8: \(y=\frac{1}{8}(x - 7)^2\). We can expand \((x - 7)^2=x^{2}-14x + 49\), so \(y=\frac{1}{8}x^{2}-\frac{14}{8}x+\frac{49}{8}=\frac{1}{8}x^{2}-\frac{7}{4}x+\frac{49}{8}\)

Answer:

\(\frac{1}{8}(x - 7)^2\) (or \(\frac{1}{8}x^{2}-\frac{7}{4}x+\frac{49}{8}\))