Coordinate Geometry Long Questions (Question 5)


Question 5:
In the diagram, the equation of FMG is y = – 4. A point P moves such that its distance from E is always half of the distance of E from the straight line FG. Find
(a) The equation of the locus of P,
(b) The x-coordinate of the point of intersection of the locus and the x-axis.

Solution:
(a) 
Gradient of the straight line FMG = 0
EM is perpendicular to FMG, so gradient of EM also = 0, equation of EM is x = 2
Thus, coordinates of point M = (2, 4).

Let coordinates of point P= (x, y).
Given PE = ½ EM
2PE = EM
2 [(x – 2)2+ (y – 4)2]½ = [(2 2)2 + (4 (4))2]½
4 (x2 – 4x + 4 + y2 – 8y +16) = (0 + 64) → (square for both sides)
4x2 – 16x + 16 + 4y2 – 32y + 64 = 64
4x2 + 4y2 – 16x – 32y + 16 = 0
x2 + y2 – 4x – 8y + 4 = 0

(b) 
x2 + y2 – 4x – 8y + 4 = 0
At x axis, y = 0.
x2 + 0 – 4x – 8(0) + 4 = 0
x2  – 4x+ 4 = 0
(x – 2) (x – 2) = 0
x = 2

The x-coordinate of the point of intersection of the locus and the x-axis is 2.

Leave a Reply

Your email address will not be published. Required fields are marked *