S4 Mathematics Final Examination

Detailed Solutions - Questions 16 & 17

Wednesday, June 24, 2026

Question 16

A circle passes through the origin $O(0,0)$ and meets the $y$ -axis and $x$ -axis at $A(0,12)$ and $B(5,0)$ respectively. A straight line $L$ parallel to $AB$ cuts the circle at $C$ and $D$ in quadrant I.

(a) Equation of AB

Slope $m_{AB} = \dfrac{0-12}{5-0} = -\dfrac{12}{5}$ ; with the intercept form:

\frac{x}{5} + \frac{y}{12} = 1 \;\Longrightarrow\; 12x + 5y - 60 = 0

(b) Centre G and radius r

As $\angle AOB = 90^\circ$ , $AB$ is a diameter (angle in a semicircle), so $G$ is its midpoint:

G = (2.5,\, 6), \qquad r = \tfrac{1}{2}\sqrt{5^2+12^2} = 6.5

d^2 = r^2 - \left(\tfrac{CD}{2}\right)^2 = 6.5^2 - 2.5^2 = 36 \;\Longrightarrow\; d = 6

(d) Point M on CD closest to G

Step $d=6$ from $G$ along the unit normal $\tfrac{1}{13}(12,5)$ :

M = G + \frac{6}{13}(12,5) = \left(\frac{209}{26}, \frac{108}{13}\right) \approx (8.04,\, 8.31)

Diameter AB, parallel chord CD, and closest point M at distance d = 6 from G.

Question 17

f(x) = \frac{1}{k}\left[x^2 + (2k+8)x + (25k-20)\right]

(a)(i) Show f passes through F(2, 29)

f(2) = \frac{1}{k}\left[4 + 2(2k+8) + 25k - 20\right] = \frac{29k}{k} = 29 \;\checkmark

(a)(ii) Vertex of f(x)

\text{Vertex} = \left(-k-4,\; 17 - k - \frac{36}{k}\right)

(b)(i) Vertex U of g(x) = f(-x) + 8

U = \left(k+4,\; 25 - k - \frac{36}{k}\right)

(b)(ii) Values of k giving minimum value 12

$25 - k - \tfrac{36}{k} = 12 \Rightarrow k^2 - 13k + 36 = 0$ :

(k-4)(k-9) = 0 \;\Longrightarrow\; k = 4 \ \text{or}\ k = 9

(b)(iii) Is the angle claim true?

For $k=4$ , $\angle FOV \approx 29.7^\circ$ . Since $29.7^\circ < 32^\circ$ , the claim is FALSE.