The equation of conic that satisfies the given conditions parabola, focus (5, −2), vertex (5, 6) is x^2 -10x -16y + 9 = 0.
According to the given question.
The focus of parabola is (5, -2).
The vertex of the parabola is (5, 6).
Points (5, -2) and (5, 6) lie on line x = 5.
So, the axis of the parabola is the line x = 5.
Focus of the parabola lies below the vertex.
So, the parabola is downwards opening.
So, the equation for the conic is (x - 5)^2 = -4a(y + 1).
Distance between the focus and vertex,a = -6-(-2) = -4
Therefore, the equation of conic that satisfies the given conditions parabola, focus (5, −2), vertex (5, 6) is given by
(x - 5)^2 = -4(-4)(y + 1).
⇒ x^2 + 25 -10x = 16(y + 1)
⇒ x^2 + 25 -10x = 16y + 16
⇒ x^2 + 25 -10x -16y -16=0
⇒ x^2 -10x -16y + 9 = 0.
Hence, the equation of conic that satisfies the given conditions parabola, focus (5, −2), vertex (5, 6) is x^2 -10x -16y + 9 = 0.
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