# Find the eq

Given that we need to find the equation of the parabola whose focus is (5, 2) and having a vertex at (3, 2).

We know that the directrix is perpendicular to the axis and vertex is the midpoint of focus and the intersection point of axis and directrix.

Let us find the slope of the axis. We know that the slope of the straight line passing through the points (x1, y1) and (x2, y2) is .

m1 = 0.

Here the axis is parallel to the x-axis, so the directrix should be parallel to the y-axis.

Let us assume the intersection point on directrix is (x1, y1).

x + 5 = 6 and y + 2 = 4

x = 1 and y = 2.

The point on directrix is (1, 2).

We know that the equation parallel to y-axis is x = k. So, the equation of the directrix is x = 1.

Let P(x, y) be any point on the parabola.

We know that the point on the parabola is equidistant from focus and directrix.

We know that the distance between two points (x1, y1) and (x2, y2) is .

We know that the perpendicular distance from a point (x1, y1) to the line ax + by + c = 0 is .

SP = PM

SP2 = PM2

x2 - 10x + y2 - 4y + 29 = x2 - 2x + 1

y2 - 8x - 4y + 28 = 0

The equation of the parabola is y2 - 8x - 4y + 28 = 0.

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