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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,
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).
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