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A point moves in such a way that it is always the same distance from the point F(-3, -7) as it is from the line y = -1. Determine the equation of the locus in standard form.
This is the first time I come across a question in this form. How do I make use of the F(-3, -7) point and y = -1 line to write the equation?
Thanks.
Hello, shenton!
Quote:
A point moves in such a way that it is always the same distance
from the point F(-3,-7) as it is from the line y = -1.
Determine the equation of the locus in standard form.
It might help you make a sketch . . .
Let P(x,y) be any point satisfying the restriction.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .________________
The distance from F(-3,-7) to P(x,y) is: .FP .= .√(x + 3)² + (y + 7)²
The distance from P(x,y) to the line y = -1 is: .y + 1
. . . . . . . . . . .________________
So we have: .√(x + 3)² + (y + 7)² .= .y + 1
Square: .x² + 6x + 9 + y² + 14y + 49 .= .y² + 2y + 1
. . x² + 6x + 9 .= .-12y - 48
. . (x + 3)² .= .-12(y + 4)
This is a down-opening parabola with vertex (-3, -4).