# equation of locus

• Mar 11th 2007, 05:06 PM
shenton
equation of locus
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.
• Mar 11th 2007, 08:39 PM
Soroban
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.

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).