# What is the equation of the quadratic graph with a focus of (5, -1) and a directrix of y = 1?

**Solution:**

Let P(x, y) be the moving point. A quadratic graph is that of a parabola. The parabola is the locus of a point P which moves such that the distance of the point from focus and the directrix is equal. Here it is given that the focus is S(5, -1) and the directrix y = k= 1. Draw PM perpendicular to y = k = 1 then, coordinates of M(x, 1).

By definition and the diagram,

PS = PM

Squaring both the sides,

PS^{2} = PM^{2}

(x-5)^{2}+ (y+1)^{2}= (x-x)^{2}+ (y-1)^{2}(using the distance formula between two points)

x^{2} -10x + 25 + y^{2} +2y + 1 = y^{2} -2y + 1

x^{2} -10x + 25 = -4y

(x-5)^{2} = -4(y-0), which is of the form (x- h)^{2}= 4a (y - k).

## What is the equation of the quadratic graph with a focus of (5, -1) and a directrix of y = 1?

**Summary: **

The equation of the quadratic graph with a focus of (5,-1) and a directrix of y=1 is (x-5)^{2} = -4(y-0).

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