**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*2

*PE*=*EM*2 [(

*x –*2)^{2}+ (*y*– 4)^{2}]^{½}= [(2*–*2)^{2}+ (4*–*(*–*4))^{2}]^{½}4 (

*x*^{2}– 4*x*+ 4 +*y*^{2}– 8*y*+16) = (0 + 64) → (square for both sides)4

*x*^{2}– 16*x*+ 16 + 4*y*^{2}– 32*y*+ 64 = 644

*x*^{2}+ 4*y*^{2}– 16*x*– 32*y*+ 16 = 0*x*

^{2}+

*y*

^{2}– 4

*x*– 8

*y*+ 4 = 0

**(b)**

*x*

^{2}+

*y*

^{2}– 4

*x*– 8

*y*+ 4 = 0

At

*x*axis,*y*= 0.*x*

^{2}+ 0 – 4

*x*– 8(0) + 4 = 0

*x*

^{2}– 4

*x*+ 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.