# find the equation of the parabola with axis parallel to Ox, latus rectum 1, and passing  through (3,1),(-5,5) analytic geometry conic sections the parabola

Let the vertex of the parabola be (x0, y0). As the latus rectum is 1, a = 1/4. The equation of the parabola is (y - y0)^2 = (x - x0)

It passes through (3,1) and (-5,5)

=> (5 - y0)^2 = (- 5 - x0) and (1 - y0)^2...

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Let the vertex of the parabola be (x0, y0). As the latus rectum is 1, a = 1/4. The equation of the parabola is (y - y0)^2 = (x - x0)

It passes through (3,1) and (-5,5)

=> (5 - y0)^2 = (- 5 - x0) and (1 - y0)^2 = ( 3 - x0)

x0 = -5 - (5 - y0)^2

substituting in (1 - y0)^2 = ( 3 - x0)

=> (1 - y0)^2 = 3 + 5 + (5 - y0)^2

=> (1 - y0)^2 - (5 - y0)^2 = 8

=> (1- y0 - 5 + y0)(1 - y0 + 5 - y0) = 8

=> -4*(6 - 2*y0) = 8

=> 2y0 - 6 = 2

=> 2y0 = 8

=> y0 = 4

x0 = -5 - (5 - y0)^2

=> x0 = -5 - ( 5 - 4)^2

=> x0 = -5 - 1

=> x0 = -6

The equation of the parabola is (y - 4)^2 = x + 6

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