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Finding the vertex focus directrix and axis of symmetry of a parabola

Finding the vertex focus directrix and axis of symmetry of a parabola-example-1

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Equation:


(y+1)^2=6(x-5)

The vertex is given by the following formula:


(y-k)^2=4p(x-h)

where the vertex is (h, k). Thus, in our equation k = -1 and h = 5, and the vertex

is (5, -1).

Additionally, the focus is given by (h+p, k). In our case:


p=(6)/(4)=(3)/(2)

Then, the focus is:


(5+(3)/(2),-1)

Simplifying:


((13)/(2),-1)

The directrix is x = h - p:


x=5-(3)/(2)=(7)/(2)

Finally, the axis of symmetry is y = -1.

User Anjali Bhimani
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