S4 Final Examination - Mathematics

Questions 16 & 17, worked solutions

Coordinate geometry of a circle and the analysis of a parameterised quadratic, with every step rendered cleanly.

Q16Circle through O, A and B

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

(a)Equation of AB

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

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

(b)Centre G and radius r

Since $\angle AOB = 90^\circ$ and the circle passes through $O, A, B$ , the chord $AB$ is a diameter (angle in a semicircle). So $G$ is the midpoint of $AB$ :

G = \left(\tfrac{0+5}{2}, \tfrac{12+0}{2}\right) = (2.5,\, 6), \qquad r = \tfrac{1}{2}\sqrt{5^2 + 12^2} = 6.5

(c)Shortest distance between AB and CD when CD = 5

The perpendicular distance from the centre to a chord of length $\ell$ is $\sqrt{r^2 - (\ell/2)^2}$ . As $AB$ is a diameter through $G$ , the gap between the parallel lines equals that distance:

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

$M$ lies a distance $d = 6$ from $G$ along the unit normal to $AB$ , which is $\tfrac{1}{13}(12, 5)$ :

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 the closest point

M

at distance

d = 6

from

G

Q17A parameterised quadratic

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 + (2k+8)(2) + 25k - 20\right] = \frac{29k}{k} = 29 \;\checkmark

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

Completing the square, the axis of symmetry is at $x = -(k+4)$ , giving:

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

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

Reflecting in the $y$ -axis and shifting up by 8 maps the vertex to:

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

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

Setting the minimum $25 - k - \tfrac{36}{k} = 12$ gives $k^2 - 13k + 36 = 0$ , so:

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

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

For $k = 4$ , the relevant angle is $\angle FOV \approx 29.7^\circ$ . The claim that the angle exceeds $32^\circ$ is therefore:

False- since

29.7^\circ < 32^\circ

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