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,

PS² = PM²

(x - 3)²+ (y - 1)²= (x - x)²+ (y - 5)²

x² -6x + 9 + y² - 2y + 1 = y² -10y + 25

x² -6x + 9 - 2y + 1 = -10y + 25

x² -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).