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Define the axis of symmetry, vertex, focus and directrix for

16(y-4)= (x-3)^2

NO LINKS!!


1 Answer

7 votes

Answer:

see attached

Explanation:

The equation is in the form ...

4p(y -k) = (x -h)^2 . . . . . (h, k) is the vertex; p is the focus-vertex distance

Comparing this to your equation, we see ...

p = 4, (h, k) = (3, 4)

p > 0, so the parabola opens upward. The vertex is on the axis of symmetry. That axis has the equation x=x-coordinate of vertex. This tells you ...

vertex: (3, 4)

axis of symmetry: x = 3

focus: (3, 8) . . . . . 4 units up from vertex

directrix: y = 0 . . . horizontal line 4 units down from vertex

Define the axis of symmetry, vertex, focus and directrix for 16(y-4)= (x-3)^2 NO LINKS-example-1
User Palanikumar
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