Find the vertex, focus, directrix, and focal width of the parabola. x= 4y^2?
Could you please help me find the vertex, focus, directrix, and focal width of the parabola defined by the equation ( x = 4y^2 )? I would appreciate a detailed explanation of each component and how they relate to the parabola's geometry. Thank you!
1 Answers
Jan 26, 2025
given the horizontal 4p(x − h) = (y − k)^2
the vertex is at (h,k)
the focus is at (h+p,k)
the directrix is at x= h−p
and the focal width = |4p|
so x = 4y^2
∴ (1/4)(x − 0) = (y − 0)^2
∴ 4(1/16)(x − 0) = (y − 0)^2
so the vertex is at (h,k) = (0,0)
the focus is at (h+p,k) = (1/16,0)
the directrix is x = h−p ⇒ x = -1/16
the focal width is |4p| = 1/4
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