Find the equation of the parabola whose axis parallel to y-axis and which passes through the points (1,0) , (0,0)

and (-2,4) .

Rates assured

The eqn. of the parabola is (x-1/2)²=(3/2)(y+1/6).

Since the axes is parallel to Y-axis,It's eqn. is of the form

(x-h)²=4a(y-k).Now,Substitute the given points in the eqn. to find a,h,k.

h²=-4ak  ,  (h-1)²=-4ak  ,  (h+2)²=4a(4-k)

Subtracting 1 from 2,we get,h=1/2.So,ak=-1/16 and a(4-k)=25/16

Dividing these two,we get,k=-1/6 and hence,a=3/8.

Thus,the eqn. is (x-1/2)²=(3/2)(y+1/6).

On simplifying,it becomes 2x²-2x-3y=0.

General form of Parabola with Axis parallel to y axis: (x-a)² = 4k(y-b)

From (0,0), we have a² = -4kb                                                             ----------------------------         A

From (1,0), we have 1+a²-2a = -4kb                                                    ------------------------------      B

From (-2,4), we have 4+a²+4a = 16k - 4kb                                        --------------------------------       C

From A and B,  1-2a = 0, so a = 1/2.

From C and A, 4+4a=16k or 1+a=4k or k=1/4 + a/4 = 3/8. so k = 3/8.

From A, 1/4 = -3b/2  or b = -1/6.

So equation of the parabola is (x - 1/2)² = 4 * (3/8) * (y + 1/6).

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