# find the equation of the parabola with vertex on the line y=x, axis parallel to Ox, and passing through (6,-2),(3,4).analytic geometry,. conic sections the parabola

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### 1 Answer

The general equation of a parabola with an axis parallel to Ox is (y - y0)^2 = 4a(x - x0), where the vertex is (x0, y0).

Here the vertex lies on the line y = x, so we can write it as

(y - c)^2 = 4a(x - c)

The parabola goes through the points (6 , -2) and (3 , 4). We get two equations to solve for a and c.

(4 - c)^2 = 4a( 3 - c) and ( -2 - c)^2 = 4a(6 - c)

(4 - c)^2 = 4a( 3 - c)

=> 16 + c^2 - 8c = 12a - 4ac ...(1)

( -2 - c)^2 = 4a(6 - c)

=> 4 + c^2 + 4c = 24a - 4ac ...(2)

(1) - (2)

=> 12 - 12c = -12a

=> c - 1 = a

=> c = a + 1

Substituting in (2)

=> 4 + (a + 1)^2 + 4(a + 1) = 24a - 4a(a + 1)

=> 4 + a^2 + 1 + 2a + 4a + 4 = 24a - 4a^2 - 4a

=> 5a^2 - 14a + 9 = 0

=> 5a^2 - 9a - 5a + 9 = 0

=> a(5a - 9) - 1(5a - 9) = 0

=> (a - 1)(5a - 9)

=> a = 1 and a = 9/5

c = 2 and c = 14/5

**The equation of the parabola is (y - 2)^2 = 4(x - 2) and (y - 14/5)^2 = (36/5)(x - 14/5)**