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for the parabola: $(x - 2)^2 = 8(y - 1)$ find the vertex and focus. mak…

Question

for the parabola: $(x - 2)^2 = 8(y - 1)$
find the vertex and focus. make sure to enter your answers as coordinates.
vertex:
focus:
write the equation of the directrix, and then sketch the parabola, the directrix and place a dot on the vertex and the focus.
directrix
graphig tool tutorial +

Explanation:

Step1: Recall the standard form of a parabola

The standard form of a parabola that opens up or down is \((x - h)^2=4p(y - k)\), where \((h,k)\) is the vertex. For the given equation \((x - 2)^2=8(y - 1)\), we can compare it with \((x - h)^2=4p(y - k)\).
Here, \(h = 2\), \(k = 1\). So the vertex \((h,k)=(2,1)\).

Step2: Find the value of \(p\)

Since \((x - 2)^2=8(y - 1)\) and \((x - h)^2=4p(y - k)\), then \(4p=8\), so \(p = 2\).

Step3: Find the focus

The focus of a parabola \((x - h)^2=4p(y - k)\) is \((h,k + p)\). Substituting \(h = 2\), \(k = 1\), \(p = 2\), we get \((2,1 + 2)=(2,3)\).

Step4: Find the directrix

The equation of the directrix of a parabola \((x - h)^2=4p(y - k)\) is \(y=k - p\). Substituting \(k = 1\), \(p = 2\), we get \(y=1-2=-1\).

Answer:

Vertex: \((2,1)\)
Focus: \((2,3)\)
Directrix: \(y=-1\)