# Find the standard form of the equation of the parabola with the given characteristics. Vertex: (4, 3); focus: (6, 3)

**Solution:**

Since the vertex is at V(4, 3) and the focus F is at (6, 3) the directrix is at x = 2.

We know that a point P(x, y) lies on the parabola only and only if its distance from the focus is the same as its distance from the directrix which is x = 2.

In other words the distance of point P(x, y) on the parabola from the point on the directrix i.e.(2, y) is the same as its distance from the focus F(6,3). Therefore we can write:

\(\sqrt{(x-6)^{2} + (y-3)^{2}} = \sqrt{(x-2)^{2} + (y-y)^{2}}\)

Squaring both sides we have:

\((x-6)^{2} +(y-3)^{2} = (x-2)^{2} + (0)^{2}\)

\((y-3)^{2} = (x-2)^{2} - (x-6)^{2}\)

We know that

a^{2} - b^{2} = (a - b) (a + b) which implies

(y - 3)^{2} = (x - 2 + x - 6)(x - 2 -x + 6)

(y - 3)^{2} = (2x - 8)(4)

2x - 8 = (y - 3)^{2} / 4

2x = (y - 3)^{2} / 4 + 8

x = (y - 3)^{2} / 8 + 4.

## Find the standard form of the equation of the parabola with the given characteristics. Vertex: (4, 3); focus: (6, 3)

**Summary:**

The standard form of the equation of the parabola with the given characteristics i.e.Vertex: (4, 3); focus: (6, 3) is x = (y - 3)² / 8 + 4

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