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# What is the equation of the quadratic graph with a focus of (3,1) and a directrix of y=5?

**Solution:**

Let P(x, y) be the moving point. It is the locus of a point P which moves such that the distance of the point from focus S(3, 1) and the directrix y= k = 5 is equal. Draw PM perpendicular to y = k = 5, then coordinates of M(x , 5)

By definition and the diagram,

PS = PM

Squaring both the sides,

PS^{2} = PM^{2}

(x - 3)^{2}+ (y - 1)^{2}= (x - x)^{2}+ (y - 5)^{2}

x^{2} -6x + 9 + y^{2} - 2y + 1 = y^{2} -10y + 25

x^{2} -6x + 9 - 2y + 1 = -10y + 25

x^{2} -6x + 9 = -8y + 24

(x - 3)^{2}= -8y + 24

(x - 3)^{2}= -8(y - 3) , which is of the form (x - h)^{2}= 4a (y - k).

## What is the equation of the quadratic graph with a focus of (3,1) and a directrix of y=5?

**Summary: **

The equation of the quadratic graph with a focus of (3, 1) and a directrix of y = 5 is (x - 3)^{2}= -8(y - 3).

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