2022 H2 Mathematics Paper 1 Question 5

Equations and Inequalities

Answers

(b)

A(−45,725),{A \left( - \frac{4}{5}, \frac{72}{5} \right),} B(−12,8).{B \left( - 12, 8 \right).}

Full solutions

(a)

(x+8)2+(mx−14)2=52x2+16x+64+m2x2−28mx+196=52\begin{gather*} (x + 8)^2 + (mx - 14)^2 = 52 \\ x^2 + 16 x + 64 + m^2 x^2 - 28mx + 196 = 52 \end{gather*}
(1)(m2+1)x2+(−28m+16)x+208=0(1) \qquad (m^2 + 1)x^2 + (- 28 m + 16)x + 208 = 0
Since the line is a tangent to the curve,
b2−4ac=0(−28m+16)2−4(m2+1)(208)=0−48m2−896m−576=03m2+56m+36=0  ■\begin{gather*} b^2 - 4ac = 0 \\ (- 28 m + 16)^2 - 4(m^2 + 1)(208) = 0 \\ - 48 m^2 - 896 m - 576 = 0 \\ 3 m^2 + 56 m + 36 = 0 \; \blacksquare \end{gather*}

(b)

(m+18)(3m+2)=0m=−18 or m=−23\begin{gather*} (m + 18)(3 m + 2) = 0 \\ m = - 18 \quad \textrm{ or } \quad m = - \frac{2}{3} \end{gather*}
Substituting m=−18{m=- 18} into (1),{(1),}
325x2+520x+208=025x2+40x+16=0(5x+4)2=0\begin{gather*} 325 x^2 + 520 x + 208 = 0 \\ 25 x^2 + 40 x + 16 = 0 \\ (5 x + 4)^2 = 0 \end{gather*}
x=−45y=−18(−45)=725\begin{align*} x &= {\textstyle - \frac{4}{5}} \\ y &= {\textstyle - 18 \left( - \frac{4}{5} \right)} \\ &= {\textstyle \frac{72}{5}} \end{align*}
Substituting m=−23{m=- \frac{2}{3}} into (1),{(1),}
139x2+1043x+208=0x2+24x+144=0(x+12)2=0\begin{gather*} \frac{13}{9} x^2 + \frac{104}{3} x + 208 = 0 \\ x^2 + 24 x + 144 = 0 \\ (x + 12)^2 = 0 \end{gather*}
x=−12y=−23(−12)=8\begin{align*} x &= {\textstyle - 12} \\ y &= {\textstyle - \frac{2}{3} \left( - 12 \right)} \\ &= {\textstyle 8} \end{align*}
Coordinates of A{A} and B{B} are (−45,725){\left( - \frac{4}{5}, \frac{72}{5} \right)} and (−12,8).  ■{\left( - 12, 8 \right). \; \blacksquare}

Question Commentary

This question tests our ability to work algebraically: part(a) uses our prior knowledge of discriminant, while part (b) involves systematically solving for x{x} and y{y} after finding the values of m.{m.}