F4 A.math

1. Let the coordinates of P be (xp, yp)
M is the mid-point of PQ, so that
x-coordiates M: xm = (2+xp)/2
xp = 2xm - 2

y-coordiates M: ym = (3+yp)/2
yp = 2ym - 3

P (xp, yp) is a point on the curve x2 + y2 =25, so that

(xp)2 + (yp)2 =25

(2xm - 2)2 + (2ym - 3)2 = 25

4(xm)2 - 8(xm) + 4 + 4(ym)2 - 12(ym) + 9 = 25

4(xm)2 - 8(xm) + 4(ym)2 - 12(ym) = 12

(xm)2 - 2(xm) + (ym)2 - 3(ym) = 3

therefore, the equation of locus of M is x2 - 2x + y2 - 3y = 3
2. Let the coordinates of P be (xp, yp)
PA2 - PB2 = k
{√[(yp - 1)2 + [xp -(-3)]2]}2 - {√[(yp - 4)2 + (xp -3)2]}2 = k

[(yp - 1)2 + [xp -(-3)]2] - [(yp - 4)2 + (xp -3)2]= k

[(yp)2 - 2yp + 1 + (xp)2 + 6xp+ 9] - [(yp)2 - 8yp + 16 + (xp)2 - 6xp +9] = k

6yp + 12xp - 15= k

therefore, the equation of locus of P is 12x + 6y - (k+15) = 0

b) equation of locus of P is 12x + 6y - (k+15) = 0
y = [-12/6]x + [(k+15)/6]
slope of the equation of locus of P = -2

slope of AB = (4 - 1)/ [3-(-3)] = 3 /6 = 1/2
slope of AB x slope of equation of locus of P= (1/2)(-2) = -1
therefore, the above locus is a straight line perpendicular to AB

c) Substititue B (3, 4) into the equation of locus of P: 12x + 6y - (k+15) = 0
12(3) + 6(4) - (k +15) = 0
60 = k + 15
k= 45

Good !!!!!!!!!!!!!!!!!!