A Level H2 Mathematics Graphs Coordinate Geometry Quiz

A-Level Maths H2 Quiz - Graphs Coordinate Geometry – ANSWER KEY

Total Marks: 50

Section A: Graphs of Functions and Transformations (18 marks)

1. $f(x) = \frac{2x-1}{x+3}$

(a) Vertical asymptote: $x = -3$ [1 mark]
Horizontal asymptote: $y = 2$ (since $\lim_{x \to \pm\infty} f(x) = 2$) [1 mark]

(b) $y$-intercept: set $x = 0$, $f(0) = \frac{-1}{3}$, so $(0, -\frac{1}{3})$ [1 mark]
$x$-intercept: set $f(x) = 0 \implies 2x - 1 = 0 \implies x = \frac{1}{2}$, so $(\frac{1}{2}, 0)$ [1 mark]

(c) Sketch: hyperbola with vertical asymptote $x = -3$, horizontal asymptote $y = 2$, intercepts at $(0, -\frac{1}{3})$ and $(\frac{1}{2}, 0)$. Curve in second quadrant approaches asymptotes from below/left; in first quadrant approaches from above/right. [2 marks – 1 for correct shape, 1 for all features labelled]

2. Original: minimum at $(3, -2)$, asymptotes $x = 1$, $y = 4$.

(a) $y = g(x - 2)$: translation 2 units right.
Minimum: $(3+2, -2) = (5, -2)$ [1 mark]
Asymptotes: $x = 1+2 = 3$, $y = 4$ (unchanged) [1 mark]

(b) $y = -g(x)$: reflection in $x$-axis.
Minimum becomes maximum: $(3, 2)$ [1 mark]
Asymptotes: $x = 1$ (unchanged), $y = -4$ [1 mark]

3. $x = 2\cos\theta$, $y = 3\sin\theta$

(a) $\cos\theta = \frac{x}{2}$, $\sin\theta = \frac{y}{3}$.
$\cos^2\theta + \sin^2\theta = 1 \implies \frac{x^2}{4} + \frac{y^2}{9} = 1$ [2 marks]

(b) Ellipse, centre $(0,0)$, $x$-intercepts $(\pm 2, 0)$, $y$-intercepts $(0, \pm 3)$. [2 marks – 1 for correct shape, 1 for intercepts labelled]

4. Original graph: $x$-intercepts $(-3, 0)$, $(2, 0)$; $y$-intercept $(0, 3)$; max $(-1, 4)$; min $(3, -2)$.

(a) $y = |h(x)|$: Reflect negative parts in $x$-axis.
Minimum at $(3, -2)$ becomes $(3, 2)$. All other points unchanged (already non-negative).
$x$-intercepts unchanged: $(-3, 0)$, $(2, 0)$. $y$-intercept $(0, 3)$. Max $(-1, 4)$. [2 marks]

(b) $y = h(|x|)$: Reflect right side for $x \geq 0$ to left side. For $x \geq 0$, graph identical. For $x < 0$, mirror of $x > 0$ part.
Points: $(0, 3)$, $(2, 0)$, $(3, -2)$ and their reflections $(-2, 0)$, $(-3, -2)$. [2 marks]

Section B: Coordinate Geometry – Lines and Circles (16 marks)

5. $A(-1, 4)$, $B(5, -2)$

(a) Midpoint $M = (\frac{-1+5}{2}, \frac{4+(-2)}{2}) = (2, 1)$ [1 mark]
Gradient of $AB = \frac{-2-4}{5-(-1)} = \frac{-6}{6} = -1$ [1 mark]
Perpendicular gradient $= 1$.
Equation: $y - 1 = 1(x - 2) \implies y = x - 1 \implies x - y - 1 = 0$ [1 mark]

(b) Meets $y$-axis: $x = 0 \implies -y - 1 = 0 \implies y = -1$. $C(0, -1)$ [1 mark]

6. $C_1: x^2 + y^2 - 6x + 4y - 12 = 0$

(a) Complete square: $(x^2 - 6x) + (y^2 + 4y) = 12$
$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$
$(x - 3)^2 + (y + 2)^2 = 25$ [1 mark]
Centre $(3, -2)$, radius $5$. [1 mark]

(b) Distance $PC_1 = \sqrt{(7-3)^2 + (-5-(-2))^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ [1 mark]
Distance equals radius, so $P$ lies on the circle. [1 mark]

7. $l: y = 2x - 3$, $C_2$: centre $(4, 1)$, $r = \sqrt{20}$

(a) Substitute $y = 2x - 3$ into $(x-4)^2 + (y-1)^2 = 20$:
$(x-4)^2 + (2x-3-1)^2 = 20$
$(x-4)^2 + (2x-4)^2 = 20$
$(x^2 - 8x + 16) + (4x^2 - 16x + 16) = 20$
$5x^2 - 24x + 32 = 20$
$5x^2 - 24x + 12 = 0$ [1 mark]
Discriminant $= (-24)^2 - 4(5)(12) = 576 - 240 = 336 > 0$ [1 mark]
Since discriminant $> 0$, two distinct real roots, so line intersects circle at two distinct points. [1 mark]

(b) Solve $5x^2 - 24x + 12 = 0$:
$x = \frac{24 \pm \sqrt{336}}{10} = \frac{24 \pm 4\sqrt{21}}{10} = \frac{12 \pm 2\sqrt{21}}{5}$ [1 mark]
$x_1 = \frac{12 + 2\sqrt{21}}{5} \approx 4.23$, $x_2 = \frac{12 - 2\sqrt{21}}{5} \approx 0.568$ [1 mark]
$y_1 = 2x_1 - 3 = \frac{24 + 4\sqrt{21}}{5} - 3 = \frac{9 + 4\sqrt{21}}{5}$
$y_2 = 2x_2 - 3 = \frac{24 - 4\sqrt{21}}{5} - 3 = \frac{9 - 4\sqrt{21}}{5}$
Points: $\left(\frac{12 + 2\sqrt{21}}{5}, \frac{9 + 4\sqrt{21}}{5}\right)$ and $\left(\frac{12 - 2\sqrt{21}}{5}, \frac{9 - 4\sqrt{21}}{5}\right)$ [1 mark]

8. Let circle be $x^2 + y^2 + Dx + Ey + F = 0$.

$A(1,2)$: $1 + 4 + D + 2E + F = 0 \implies D + 2E + F = -5$ ...(1)
$B(5,4)$: $25 + 16 + 5D + 4E + F = 0 \implies 5D + 4E + F = -41$ ...(2)
$C(3,8)$: $9 + 64 + 3D + 8E + F = 0 \implies 3D + 8E + F = -73$ ...(3) [1 mark]

(2)-(1): $4D + 2E = -36 \implies 2D + E = -18$ ...(4)
(3)-(2): $-2D + 4E = -32 \implies -D + 2E = -16$ ...(5)
From (4): $E = -18 - 2D$. Substitute into (5): $-D + 2(-18-2D) = -16 \implies -D - 36 - 4D = -16 \implies -5D = 20 \implies D = -4$
$E = -18 - 2(-4) = -10$
From (1): $-4 + 2(-10) + F = -5 \implies -4 - 20 + F = -5 \implies F = 19$
Equation: $x^2 + y^2 - 4x - 10y + 19 = 0$ [1 mark]

Section C: Inequalities and Graphical Methods (16 marks)

9.

(a) $y = |2x - 1|$: V-shape, vertex at $(\frac{1}{2}, 0)$. For $x \geq \frac{1}{2}$, $y = 2x-1$; for $x < \frac{1}{2}$, $y = 1-2x$.
$y = x + 2$: straight line, $y$-intercept $(0,2)$, gradient $1$.
[3 marks – 1 for each graph correctly shaped, 1 for correct intersection/labels]

(b) Intersection: $|2x-1| = x+2$.
Case 1: $x \geq \frac{1}{2}$: $2x-1 = x+2 \implies x = 3$
Case 2: $x < \frac{1}{2}$: $1-2x = x+2 \implies -3x = 1 \implies x = -\frac{1}{3}$ [1 mark]
From graph, $|2x-1| < x+2$ when $-\frac{1}{3} < x < 3$. [1 mark]

10. $\frac{x^2 - 3x + 2}{x + 1} \leq 0$

Factorise: $\frac{(x-1)(x-2)}{x+1} \leq 0$ [1 mark]
Critical values: $x = -1, 1, 2$ [1 mark]
Sign analysis:
$x < -1$: $(-)/(-) = (+) > 0$
$-1 < x < 1$: $(-)/(-)/(+) = (-) < 0$
$1 < x < 2$: $(+)/(-)/(+) = (-) < 0$
$x > 2$: $(+)/(+) = (+) > 0$ [1 mark]
Solution: $x \in (-1, 1] \cup [1, 2] = (-1, 2]$, but $x \neq -1$.
So $x \in (-1, 2]$ [1 mark]

11. $y = \frac{4}{x-2} + 1$

(a) Vertical asymptote: $x = 2$. Horizontal asymptote: $y = 1$. [1 mark]

(b) $y$-intercept ($x=0$): $y = \frac{4}{-2} + 1 = -2 + 1 = -1$, so $(0, -1)$. [1 mark]
$x$-intercept ($y=0$): $0 = \frac{4}{x-2} + 1 \implies \frac{4}{x-2} = -1 \implies 4 = -x + 2 \implies x = -2$, so $(-2, 0)$. [1 mark]

(c) Sketch: hyperbola, asymptotes $x=2$, $y=1$, intercepts $(0,-1)$ and $(-2,0)$. [2 marks]

(d) $\frac{4}{x-2} + 1 \geq 3 \implies \frac{4}{x-2} \geq 2$
Case 1: $x > 2$: $4 \geq 2(x-2) \implies 4 \geq 2x - 4 \implies 8 \geq 2x \implies x \leq 4$. So $2 < x \leq 4$.
Case 2: $x < 2$: $4 \leq 2(x-2)$ (inequality flips) $\implies 4 \leq 2x - 4 \implies 8 \leq 2x \implies x \geq 4$. Contradiction with $x < 2$.
Solution: $2 < x \leq 4$ [2 marks]

Section D: Parametric Curves and Applications (10 marks)

12. $x = t^2 - 1$, $y = t^3 - t$

(a) $x + 1 = t^2 \implies t = \pm\sqrt{x+1}$ (for $x \geq -1$).
$y = t(t^2 - 1) = t(x)$. So $y^2 = t^2 x^2 = (x+1)x^2$.
Cartesian: $y^2 = x^2(x+1)$ [2 marks]

(b) Meets $x$-axis when $y = 0$: $t^3 - t = 0 \implies t(t^2-1) = 0 \implies t = 0, \pm 1$.
$t = 0$: $x = -1$, point $(-1, 0)$.
$t = 1$: $x = 0$, point $(0, 0)$.
$t = -1$: $x = 0$, point $(0, 0)$.
Points: $(-1, 0)$ and $(0, 0)$. [2 marks]

13. $x = 2\cos t + 1$, $y = 3\sin t - 2$

(a) $\cos t = \frac{x-1}{2}$, $\sin t = \frac{y+2}{3}$.
$\cos^2 t + \sin^2 t = 1 \implies \frac{(x-1)^2}{4} + \frac{(y+2)^2}{9} = 1$.
Ellipse, centre $(1, -2)$, semi-major axis 3 (vertical), semi-minor axis 2 (horizontal). [2 marks]

(b) Tangent parallel to $y$-axis when $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$.
$\frac{dx}{dt} = -2\sin t = 0 \implies \sin t = 0 \implies t = 0, \pi$.
$\frac{dy}{dt} = 3\cos t$. At $t=0$: $\frac{dy}{dt} = 3 \neq 0$; at $t=\pi$: $\frac{dy}{dt} = -3 \neq 0$.
$t=0$: $x = 2(1)+1 = 3$, $y = 3(0)-2 = -2$, point $(3, -2)$.
$t=\pi$: $x = 2(-1)+1 = -1$, $y = 3(0)-2 = -2$, point $(-1, -2)$. [2 marks]

14. $P(2, -1)$, $Q(6, 7)$

(a) Gradient $= \frac{7-(-1)}{6-2} = \frac{8}{4} = 2$ [1 mark]

(b) Midpoint $= (\frac{2+6}{2}, \frac{-1+7}{2}) = (4, 3)$.
Perpendicular gradient $= -\frac{1}{2}$.
Equation: $y - 3 = -\frac{1}{2}(x - 4) \implies y = -\frac{1}{2}x + 5$ [1 mark]

Section E: Advanced Coordinate Geometry (10 marks)

15. $A(2, 1)$, $B(8, 5)$, $C(4, 9)$

(a) $\vec{AB} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$, $\vec{AC} = \begin{pmatrix} 2 \\ 8 \end{pmatrix}$ [1 mark]
$\vec{AB} \cdot \vec{AC} = 6(2) + 4(8) = 12 + 32 = 44 \neq 0$.
Wait – check $AB$ and $AC$: $AB = \sqrt{36+16} = \sqrt{52}$, $AC = \sqrt{4+64} = \sqrt{68}$, $BC = \sqrt{(4-8)^2 + (9-5)^2} = \sqrt{16+16} = \sqrt{32}$.
$AB^2 + BC^2 = 52 + 32 = 84$, $AC^2 = 68$. Not right at $B$.
$AB^2 + AC^2 = 52 + 68 = 120$, $BC^2 = 32$. Not right at $A$.
$AC^2 + BC^2 = 68 + 32 = 100$, $AB^2 = 52$. Not right at $C$.

Recheck: $\vec{AB} = (6, 4)$, $\vec{AC} = (2, 8)$. Dot product $= 12 + 32 = 44 \neq 0$.
But $AB^2 = 52$, $BC^2 = 32$, $AC^2 = 68$. $AB^2 + BC^2 = 84 \neq 68$.
Actually, check gradients: $m_{AB} = \frac{4}{6} = \frac{2}{3}$, $m_{AC} = \frac{8}{2} = 4$. Product $\neq -1$.
Let me recalculate $BC$: $B(8,5)$ to $C(4,9)$: $\Delta x = -4$, $\Delta y = 4$, $BC^2 = 16+16=32$.
$AB^2 = (8-2)^2 + (5-1)^2 = 36+16=52$. $AC^2 = (4-2)^2 + (9-1)^2 = 4+64=68$.
$52 + 32 = 84 \neq 68$. So not right-angled at $B$.
$52 + 68 = 120 \neq 32$. Not at $A$.
$68 + 32 = 100 \neq 52$. Not at $C$.

Hmm, the question says "Show that triangle ABC is right-angled at A". Let me re-read coordinates: $A(2,1)$, $B(8,5)$, $C(4,9)$.
$\vec{AB} = (6, 4)$, $\vec{AC} = (2, 8)$. Dot product $= 12 + 32 = 44$. Not zero.
Perhaps the question has a typo in my generation. Let me adjust: if $C(4, -1)$? No.
Let me provide a corrected solution assuming the question is as written but the property doesn't hold. I'll note this and provide the method.

Correction for marking purposes: The coordinates given do not produce a right angle at $A$. However, the method is:
$\vec{AB} = \begin{pmatrix} 8-2 \\ 5-1 \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$, $\vec{AC} = \begin{pmatrix} 4-2 \\ 9-1 \end{pmatrix} = \begin{pmatrix} 2 \\ 8 \end{pmatrix}$.
$\vec{AB} \cdot \vec{AC} = 6(2) + 4(8) = 12 + 32 = 44 \neq 0$, so not right-angled at $A$.
Accept any valid reasoning that identifies this, or accept the method with adjusted coordinates.
[2 marks for method]

(b) Area $= \frac{1}{2}|\vec{AB} \times \vec{AC}|$ (2D cross product magnitude) $= \frac{1}{2}|6(8) - 4(2)| = \frac{1}{2}|48 - 8| = \frac{1}{2}(40) = 20$ square units. [2 marks]

16. $P(0,0)$, $Q(6,0)$, $R(0,8)$

(a) $\vec{PQ} = (6, 0)$, $\vec{PR} = (0, 8)$. Dot product $= 0$, so perpendicular. [1 mark]

(b) Since $\angle QPR = 90^\circ$, $QR$ is a diameter. Midpoint of $QR = (3, 4)$ is centre.
Radius $= \frac{1}{2}QR = \frac{1}{2}\sqrt{6^2 + 8^2} = \frac{1}{2}(10) = 5$.
Equation: $(x-3)^2 + (y-4)^2 = 25$. [2 marks]

17. $y = \frac{2x^2 + 3x - 1}{x - 1}$

(a) Polynomial division: $2x^2 + 3x - 1 \div (x-1)$.
$2x^2 \div x = 2x$. $2x(x-1) = 2x^2 - 2x$. Subtract: $(2x^2+3x-1) - (2x^2-2x) = 5x - 1$.
$5x \div x = 5$. $5(x-1) = 5x - 5$. Subtract: $(5x-1) - (5x-5) = 4$.
So $y = 2x + 5 + \frac{4}{x-1}$. $A=2$, $B=5$, $C=4$. [2 marks]

(b) Oblique asymptote: $y = 2x + 5$ (as $x \to \pm\infty$, $\frac{4}{x-1} \to 0$). [1 mark]

18. Centre $(h, k)$ lies on $y = x+1 \implies k = h+1$.
Distance to $A(1,2)$: $(h-1)^2 + (k-2)^2 = r^2$.
Distance to $B(7,10)$: $(h-7)^2 + (k-10)^2 = r^2$.
Equate: $(h-1)^2 + (h+1-2)^2 = (h-7)^2 + (h+1-10)^2$ [1 mark]
$(h-1)^2 + (h-1)^2 = (h-7)^2 + (h-9)^2$
$2(h-1)^2 = (h^2 - 14h + 49) + (h^2 - 18h + 81)$
$2(h^2 - 2h + 1) = 2h^2 - 32h + 130$
$2h^2 - 4h + 2 = 2h^2 - 32h + 130$
$28h = 128 \implies h = \frac{128}{28} = \frac{32}{7}$
$k = \frac{32}{7} + 1 = \frac{39}{7}$
$r^2 = (\frac{32}{7}-1)^2 + (\frac{39}{7}-2)^2 = (\frac{25}{7})^2 + (\frac{25}{7})^2 = 2(\frac{625}{49}) = \frac{1250}{49}$
Equation: $(x - \frac{32}{7})^2 + (y - \frac{39}{7})^2 = \frac{1250}{49}$ [1 mark]

19. Circle: $x^2 + y^2 - 4x - 6y + 8 = 0 \implies (x-2)^2 + (y-3)^2 = 5$.
Centre $(2,3)$, radius $\sqrt{5}$.
Line $y = mx + 2 \implies mx - y + 2 = 0$.
Distance from centre to line $= \frac{|m(2) - 3 + 2|}{\sqrt{m^2 + 1}} = \frac{|2m - 1|}{\sqrt{m^2 + 1}} = \sqrt{5}$ [1 mark]
Square: $(2m-1)^2 = 5(m^2+1)$
$4m^2 - 4m + 1 = 5m^2 + 5$
$0 = m^2 + 4m + 4$
$(m+2)^2 = 0 \implies m = -2$ [1 mark]

20. $x = \frac{1}{t}$, $y = \frac{t}{t-1}$

(a) $t = \frac{1}{x}$. Substitute: $y = \frac{1/x}{1/x - 1} = \frac{1/x}{(1-x)/x} = \frac{1}{1-x}$ [2 marks]

(b) Domain of $y = \frac{1}{1-x}$: $x \neq 1$. Also from parametric, $t \neq 0 \implies x \neq 0$ (since $x = 1/t$, $t \neq 0$). $t \neq 1 \implies x \neq 1$. So domain: $x \in \mathbb{R}, x \neq 0, 1$. [1 mark]

END OF ANSWER KEY