HKDSE M2 Question Review (2025-01-09)

Question:

1. express (a-b) in terms of ab and (a+b).

2. Using the result of the question above, try solve the following problem:

A locus of M(a, b) passes through the origin in a 2D Cartesian Plane, satisfying the condition that M is the mid-point of (g, e^g) and (h, ln(h)). By noticing that g<h and e^g > ln(h), show that the locus is y=x.

Solution:

1. (a-b)^2 = (a+b)^2 - 4ab

|a-b|= sqrt((a+b)^2 - 4ab)

2. To be provided.

Hint: In this case, a+b=0

Further hints:

At M=(0,0),

a=(g+h)/2 = 0

and b= (e^g + ln(h)) /2 =0

Since (g, e^g), (a, b), (h, ln(h)) are collinear,

we can compute the slope k as:

k = (b-e^g)/(a-g)=(b-ln(h))/(a-h)

=(e^g-ln(h))/(g-h) --------(*)

By the given inequalities, k<0

Since the locus is perpendicular to the line,

we have locus of M: b=(-1/k)a + c

At M=(0,0), c=0

Now, refer to (*),

At M=(0,0), try to prove that k= -1

-1/k = (h-g)/(e^g-ln(h)) = -(g-h)/(e^g-ln(h))

= -sqrt((gh)/(e^g•ln(h))) = -sqrt(1/k^2) > 0

Clearly k = -1

QED