Question

find the equation of the parabola with focus $(5,\frac{9}{2})$ and vertex $(5,4)$. what is the equation for the parabola? $y = \square$ (simplify your answer.)

Explanation:

Step1: Determine the form of the parabola equation

Since the \(x -\)coordinates of the focus \((5,\frac{9}{2})\) and vertex \((5,4)\) are the same, the parabola has a vertical axis. The standard form of a parabola with a vertical axis is \(y=a(x - h)^{2}+k\), where \((h,k)\) is the vertex. Here \(h = 5\), \(k = 4\), so the equation is \(y=a(x - 5)^{2}+4\).

Step2: Find the value of \(a\)

The distance \(p\) between the vertex \((h,k)=(5,4)\) and the focus \((h,k + p)=(5,\frac{9}{2})\) is given by \(p=\frac{9}{2}-4=\frac{9 - 8}{2}=\frac{1}{2}\). For a parabola \(y=a(x - h)^{2}+k\), the relationship between \(a\) and \(p\) is \(a=\frac{1}{4p}\). Substituting \(p=\frac{1}{2}\) into \(a=\frac{1}{4p}\), we get \(a=\frac{1}{4\times\frac{1}{2}}=\frac{1}{2}\).

Step3: Write the equation of the parabola

Substitute \(a=\frac{1}{2}\), \(h = 5\), and \(k = 4\) into \(y=a(x - h)^{2}+k\). We have \(y=\frac{1}{2}(x - 5)^{2}+4\). Expand \(\frac{1}{2}(x - 5)^{2}+4=\frac{1}{2}(x^{2}-10x + 25)+4=\frac{1}{2}x^{2}-5x+\frac{25}{2}+4=\frac{1}{2}x^{2}-5x+\frac{25 + 8}{2}=\frac{1}{2}x^{2}-5x+\frac{33}{2}\).

Answer:

\(y=\frac{1}{2}(x - 5)^{2}+4\) (or \(y=\frac{1}{2}x^{2}-5x+\frac{33}{2}\))