1.Find the equation of a parabola with axis parallel to Oy and passing through (1,9),(-2,9),(21,-4). -Analytical Geometry:Conic Section

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The axis of the parabola is parallel to Oy. The equation of the parabola is of the form: (x - x0)^2 = 4a(y - y0), where (x0, y0) is the vertex.

As we have 3 points through which the parabola passes, we can create 3 equations:

  • (1,9)

(1 - x0)^2...

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The axis of the parabola is parallel to Oy. The equation of the parabola is of the form: (x - x0)^2 = 4a(y - y0), where (x0, y0) is the vertex.

As we have 3 points through which the parabola passes, we can create 3 equations:

  • (1,9)

(1 - x0)^2 = 4a(9 - y0)...(1)

  • (-2 , 9)

(-2 - x0)^2 = 4a(9 - y0)...(2)

  • (21, 4)

(21 - x0)^2 = 4a(-4 - y0)...(3)

From (1) and (2)

(1 - x0)^2 = (-2 - x0)^2

=> 1 - x0 = x0 + 2

=> x0 = -1/2

Substituting in (3)

=>21.5^2 = 4a(-4 - y0) ...(4)

Substituting in (1)

=> 1.5^2 = 4a(9 - y0) ...(5)

Solving the equations (4) and (5) for a and y0, we get

4a = -460/13 and y0 = 16677/1840

The required equation of the parabola is (x + 1/2)^2 = (-460/13)(y - 16677/1840)

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