(i)

\begin{align} \text{Gradient of } AC & = {4 - 0 \over 4 - 6} \\ & = -2 \\ \\ y & = mx + c \\ y & = -2x + c \\ \\ \text{Using } & C(6, 0), \\ 0 & = -2(6) + c \\ 0 & = -12 + c \\ 12 & = c \\ \\ \text{Eqn of } AC\text{: } \phantom{0} y & = - 2x + 12 \phantom{000} \text{ --- (1)} \\ \\ \text{Eqn of parabola: } \phantom{0}y^2 & = 4x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (-2x + 12)^2 & = 4x \\ (-2x)^2 + 2(-2x)(12) + (12)^2 & = 4x \\ 4x^2 - 48x + 144 & = 4x \\ 4x^2 - 48x - 4x + 144 & = 0 \\ 4x^2 - 52x + 144 & = 0 \\ x^2 - 13x + 36 & = 0 \\ (x - 9)(x - 4) & = 0 \\ \\ x - 9 = 0 \phantom{00}&\text{or} \phantom{00} x - 4 = 0 \\ x = 9 \phantom{00}& \phantom{or-400} x = 4 \text{ (Point } A) \\ \\ \text{Substitute } & x = 9 \text{ into (1),} \\ y & = -2(9) + 12 \\ & = -6 \\ \\ \therefore & \phantom{0} B(9, - 6) \end{align}

(ii) The distance between two points can be found by $ \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } $

\begin{align} \text{Length of } AC & = \sqrt{ (4 - 6)^2 + (4 - 0)^2 } \\ & = \sqrt{20} \\ & = \sqrt{4}\sqrt{5} \\ & = 2\sqrt{5} \text{ units} \\ \\ \text{Length of } CB & = \sqrt{ (6 - 9)^2 + [0 - (-6)]^2 } \\ & = \sqrt{45} \\ & = \sqrt{9} \sqrt{5} \\ & = 3\sqrt{5} \text{ units} \end{align}

(iii)

\begin{align} \require{cancel} AC & : CB \\ 2\cancel{\sqrt{5}} & : 3\cancel{\sqrt{5}} \\ 2 &: 3 \end{align}

\begin{align} y^2 & = x \phantom{000} \text{ --- (1)} \\ \\ y & = mx + 1 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (mx + 1)^2 & = x \\ (mx)^2 + 2(mx)(1) + 1^2 & = x \\ m^2 x^2 + 2mx - x + 1 & = mx \\ m^2 x^2 + (2m - 1)x + 1 & = 0 \phantom{00000} [a = m^2, b = 2m - 1, c = 1] \\ \\ b^2 - 4ac & > 0 \phantom{00000} [\text{Two real and distinct roots}] \\ (2m - 1)^2 - 4(m^2)(1) & > 0 \\ (2m)^2 - 2(2m)(1) + (1)^2 - 4m^2 & > 0 \\ 4m^2 - 4m + 1 - 4m^2 & > 0 \\ - 4m + 1 & > 0 \\ -4m & > - 1 \\ 4m & < 1 \\ m & < {1 \over 4} \\ \\ \text{Since } m & > 0, \\ \\ 0 < m & < {1 \over 4} \text{ (Shown)} \end{align}

(i)

Since the $x$-axis is tangent to the circle, the $y$-coordinate of the centre of the circle is $13$ \begin{align} \text{Let centre of circle be } & (a, 13). \\ \\ (x - a)^2 + (y - 13)^2 & = 13^2 \\ \\ \text{When } x = 10 \text{ and } & y = 18, \phantom{000} [\text{since circle passes through } P(10, 18)]\\ (10 - a)^2 + (18 - 13)^2 & = 13^2 \\ (10)^2 - 2(10)(a) + (a)^2 + 5^2 & = 169 \\ 100 - 20a + a^2 + 25 & = 169 \\ a^2 - 20a - 44 & = 0 \\ (a - 22)(a + 2) & = 0 \\ \\ a - 22 = 0 \phantom{00}&\text{or}\phantom{00} a + 2 = 0 \\ a = 22 \phantom{0}&\phantom{or00+2} a = - 2 \\ \\ \text{For centre } & (22, 13), \\ (x - 22)^2 + (y - 13)^2 & = 13^2 \\ (x)^2 - 2(x)(22) + (22)^2 + (y)^2 - 2(y)(13) + 13^2 & = 169 \\ x^2 - 44x + 484 + y^2 - 26y + 169 & = 169 \\ x^2 + y^2 - 44x - 26y + 484 & = 0 \\ \\ \\ \text{For centre } & (-2, 13), \\ [x - (-2)]^2 + (y - 13)^2 & = 13^2 \\ (x + 2)^2 + (y - 13)^2 & = 169 \\ (x)^2 + 2(x)(2) + (2)^2 + (y)^2 - 2(y)(13) + (13)^2 & = 169 \\ x^2 + 4x + 4 + y^2 - 26y + 169 & = 169 \\ x^2 + y^2 + 4x - 26y + 4 & = 0 \end{align}

(ii)

(i)

\begin{align} \text{Mid-point of } AB, C & = \left( {-2 + 6 \over 2}, {4 + (-2) \over 2} \right) \\ & = (2, 1) \end{align}

(ii)

\begin{align} \text{Gradient of } AB & = {4 - (-2) \over -2 - 6} \\ & = -{3 \over 4} \\ \\ \text{Gradient of } DE \times -{3 \over 4} & = - 1 \\ \text{Gradient of } DE & = -1 \div -{3 \over 4} \\ & = {4 \over 3} \\ \\ y & = mx + c \\ y & = {4 \over 3}x + c \\ \\ \text{Using } & C(2, 1), \\ 1 & = {4 \over 3}(2) + c \\ 1 & = {8 \over 3} + c \\ -{5 \over 3} & = c \\ \\ \text{Eqn of diameter } DE: \phantom{.} y & = {4 \over 3}x - {5 \over 3} \\ 3y & = 4x - 5 \\ 3y - 4x + 5 & = 0 \end{align}

(iii)

\begin{align} \text{Eqn of } DE: \phantom{.} y & = {4 \over 3}x - {5 \over 3} \phantom{000} \text{ --- (1)} \\ \\ \text{Radius of circle} & = \text{Length of } AC \\ & = \sqrt{ (-2 - 2)^2 + (4 - 1)^2} \\ & = 5 \text{ units} \\ \\ \text{Eqn of circle: } \phantom{0} & (x - 2)^2 + (y - 1)^2 = 5^2 \phantom{000} \text{ --- (2)} \end{align} \begin{align} \text{Substitute } & \text{(1) into (2),} \\ (x - 2)^2 + \left( {4 \over 3}x - {5 \over 3} - 1 \right)^2 & = 5^2 \\ (x - 2)^2 + \left( {4 \over 3}x - {8 \over 3} \right)^2 & = 25 \\ (x)^2 - 2(x)(2) + (2)^2 + \left( {4 \over 3}x \right)^2 - 2\left({4 \over 3}x\right)\left(8 \over 3\right) + \left(8 \over 3\right)^2 & = 25 \\ x^2 - 4x + 4 + {16 \over 9}x^2 - {64 \over 9}x + {64 \over 9} & = 25 \\ 9x^2 - 36x + 36 + 16x^2 - 64x + 64 & = 225 \phantom{00000} [\text{Multiply by } 9] \\ 25x^2 - 100x - 125 & = 0 \\ x^2 - 4x - 5 & = 0 \\ (x - 5)(x + 1) & = 0 \\ \\ x - 5 = 0 \phantom{00}&\text{or}\phantom{00} x + 1 = 0 \\ x = 5 \phantom{00} &\phantom{or+100} x = - 1 \\ \\ \text{Substitute } & x = 5 \text{ into (1),} \\ y & = {4 \over 3}(5) - {5 \over 3} \\ & = 5 \\ \\ \therefore & \phantom{0} D(5, 5) \\ \\ \text{Substitute } & x = - 1 \text{ into (2),} \\ y & = {4 \over 3}(-1) - {5 \over 3} \\ & = - 3 \\ \\ \therefore & \phantom{0} E(-1, -3) \end{align}

(i)

\begin{align} x^2 + y^2 + 6x - 8y & = 0 \phantom{000} \text{ --- (1)} \\ \\ y & = mx - {1 \over 3} \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ x^2 + \left(mx - {1 \over 3}\right)^2 + 6x - 8\left(mx - {1 \over 3}\right) & = 0 \\ x^2 + (mx)^2 - 2(mx)\left(1 \over 3\right) + \left(1 \over 3\right)^2 + 6x - 8mx + {8 \over 3} & = 0 \\ x^2 + m^2 x^2 - {2 \over 3}mx + {1 \over 9} + 6x - 8mx + {8 \over 3} & = 0 \\ x^2 + m^2 x^2 + 6x - {26 \over 3}mx + {25 \over 9} & = 0 \\ 9x^2 + 9m^2 x^2 + 54x - 78mx + 25 & = 0 \\ (9 + 9m^2)x^2 + (54 - 78m)x + 25 & = 0 \\ \\ [a = 9 + 9m^2, b = 54 - 78m, & \phantom{.} c = 25] \end{align} \begin{align} b^2 - 4ac & = (54 - 78m)^2 - 4(9 + 9m^2)(25) \\ & = (54)^2 - 2(54)(78m) + (78m)^2 - 900 - 900m^2 \\ & = 2916 - 8424m + 6084m^2 - 900 - 900m^2 \\ & = 6084m^2 - 900m^2 - 8424m + 2916 - 900 \\ & = 5184m^2 - 8424m + 2016 \\ & = 72(72m^2 - 117m + 28) \\ & = 72(24m - 7)(3m - 4) \end{align} \begin{align} b^2 - 4ac & > 0 \\ 72(24m - 7)(3m - 4) & > 0 \\ (24m - 7)(3m - 4) & > 0 \end{align}

$$ \therefore m < {7 \over 24} \text{ or } m > {4 \over 3} $$

(ii)

\begin{align} b^2 - 4ac & = 0 \\ 72(24m - 7)(3m - 4) & = 0 \\ (24m - 7)(3m - 4) & = 0 \\ \\ 24m - 7 = 0 \phantom{00}&\text{or}\phantom{00} 3m - 4 = 0 \\ 24m = 7 \phantom{00} &\phantom{or00-4} 3m = 4 \\ m = {7 \over 24} \phantom{.} &\phantom{or00-43} m = {4 \over 3} \end{align}

(iii)

\begin{align} b^2 - 4ac & < 0 \\ 72(24m - 7)(3m - 4) & < 0 \\ (24m - 7)(3m - 4) & < 0 \end{align}

\begin{align} \therefore {7 \over 24} < m < {4 \over 3} \end{align}

(i)

\begin{align} y^2 & = -4x \phantom{000} \text{ --- (1)} \\ \\ y & = 2x + 12 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (2x + 12)^2 & = -4x \\ (2x)^2 + 2(2x)(12) + (12)^2 & = -4x \\ 4x^2 + 48x + 144 & = - 4x \\ 4x^2 + 48x + 4x + 144 & = 0 \\ 4x^2 + 52x + 144 & = 0 \\ x^2 + 13x + 36 & = 0 \\ (x + 4)(x + 9) & = 0 \\ \\ x + 4 = 0 \phantom{000}&\text{or}\phantom{00} x + 9 = 0 \\ x = - 4 \phantom{.0} &\phantom{or00+9} x = - 9 \\ \\ \text{Substitute } & x = -4 \text{ into (2),} \\ y & = 2(-4) + 12 \\ & = 4 \\ \\ \therefore & \phantom{0} B(-4, 4) \\ \\ \text{Substitute } & x = - 9 \text{ into (1),} \\ y & = 2(-9) + 12 \\ & = -6 \\ \\ \therefore & \phantom{0} A(-9, -6) \end{align}

(ii)

\begin{align} \text{Length of } AB & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2} \\ & = \sqrt{ [-9 - (-4)]^2 + (-6 - 4)^2 } \\ & = \sqrt{ 125} \\ & = \sqrt{25} \sqrt{5} \\ & = 5\sqrt{5} \text{ units} \end{align}

(iii)

\begin{align} \text{Area of } \triangle CAB & = {1 \over 2} \left| \begin{matrix} -4 & -4 & -9 & -4 \\ -4 & 4 & -6 & -4 \end{matrix} \right| \\ & = {1 \over 2} [(-4)(4)+(-4)(-6)+(-9)(-4)] \\ & \phantom{=.} - {1 \over 2} [(-4)(-4)+(4)(-9)+(-6)(-4)] \\ & = 20 \text{ sq. units} \end{align}

(iv)

\begin{align} \text{Area of } \triangle CAB & = {1 \over 2} \times \text{Base} \times {Height} \\ 20 & = {1 \over 2} (AB) (h) \\ 20 & = {1 \over 2} (5\sqrt{5})(h) \\ 40 & = 5\sqrt{5} h \\ \\ h & = {40 \over 5\sqrt{5}} \\ & = {8 \over \sqrt{5}} \\ & = {8 \over \sqrt{5}} \times {\sqrt{5} \over \sqrt{5}} \phantom{000000} [\text{Rationalise denominator}] \\ & = {8\sqrt{5} \over 5} \\ \\ \therefore \text{Perp. distance from } C \text{ to } AB & = {8\sqrt{5} \over 5} \text{ units} \end{align}

(i)

\begin{align} y & = mx \phantom{000} \text{ --- (1)} \\ \\ x^2 + y^2 - 4mx + 3 & = 0 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + (mx)^2 - 4mx + 3 & = 0 \\ x^2 + m^2 x^2 - 4mx + 3 & = 0 \\ (1 + m^2)x^2 + (-4m)x + 3 & = 0 \\ \\ [a = 1 + m^2, b = -4m, & \phantom{.} c = 3] \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ (-4m)^2 - 4(1 + m^2)(3) & < 0 \\ 16m^2 - 12 - 12m^2 & < 0 \\ 4m^2 - 12 & < 0 \\ m^2 - 3 & < 0 \\ (m)^2 - (\sqrt{3})^2 & < 0 \\ (m + \sqrt{3})(m - \sqrt{3}) & < 0 \end{align}

\begin{align} -\sqrt{3} < & m < \sqrt{3} \\ -1.732 < & m < 1.732 \\ \\ \text{Since } m & \text{ is a positive integer,} \\ m & = 1 \end{align}

(ii)

\begin{align} \text{When } & m = 1, \\ x^2 + y^2 - 4(1)x + 3 & = 0 \\ x^2 + y^2 - 4x + 3 & = 0 \\ x^2 + y^2 + 2(-2)x + 2(0)y + 3 & = 0 \\ \\ \text{Comparing with } x^2 + y^2 & + 2gx + 2fy + c = 0, \\ g & = - 2 \\ f & = 0 \\ c & = 3 \\ \\ \therefore \text{Centre } & (2, 0) \\ \\ \text{Radius} & = \sqrt{ g^2 + f^2 - c } \\ & = \sqrt{ (-2)^2 + (0)^2 - 3 } \\ & = 1 \text{ units} \end{align}

The two possible tangents (in blue) are perpendicular to the radius of the circle (Circle property from E Maths). Note that both tangents have gradient of $1$ (since $m = 1$)

\begin{align} \text{Gradient of diameter} \times 1 & = -1 \\ \text{Gradient of diameter} & = -1 \\ \\ y & = mx + c \\ y & = -x + c \\ \\ \text{Using} & \text{ centre (2, 0)}, \\ 0 & = -(2) + c \\ 0 & = -2 + c \\ 2 & = c \\ \\ \text{Equation of diameter: } \phantom{0} y & = 2 - x \phantom{000} \text{ --- (1)} \\ \\ \text{Equation of circle: } \phantom{0} x^2 + y^2 - 4x + 3 & = 0 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + (2 - x)^2 - 4x + 3 & = 0 \\ x^2 + (2)^2 - 2(2)(x) + (x)^2 - 4x + 3 & = 0 \\ x^2 + 4 - 4x + x^2 - 4x + 3 & = 0 \\ 2x^2 - 8x + 7 & = 0 \\ \\ x & = {-b \pm \sqrt{ b^2 - 4ac } \over 2a} \\ & = { -(-8) \pm \sqrt{ (-8)^2 - 4(2)(7) } \over 2(2) } \\ & = { 8 \pm \sqrt{ 8 } \over 4} \\ & = { 8 \pm \sqrt{4} \sqrt{2} \over 4} \\ & = { 8 \pm 2\sqrt{2} \over 4} \\ & = 2 \pm {1 \over 2} \sqrt{2} \\ \\ \text{Substitute } & x = 2 + {1 \over 2}\sqrt{2} \text{ into (1),} \\ y & = 2 - \left(2 + {1 \over 2}\sqrt{2}\right) \\ & = -{1 \over 2} \sqrt{2} \\ \\ \therefore & \phantom{0} \left(2 + {1 \over 2}\sqrt{2}, -{1 \over 2}\sqrt{2} \right) \\ \\ \text{Substitute } & x = 2 - {1 \over 2}\sqrt{2} \text{ into (2),} \\ y & = 2 - \left(2 - {1 \over 2}\sqrt{2} \right) \\ & = {1 \over 2}\sqrt{2} \\ \\ \therefore & \phantom{0} \left(2 - {1 \over 2}\sqrt{2}, {1 \over 2}\sqrt{2} \right) \end{align}

\begin{align} y & = mx + c \\ y & = x + c \phantom{00000} [\text{Tangent has gradient } 1] \\ \\ \text{Using } & \left(2 + {1 \over 2}\sqrt{2}, -{1 \over 2}\sqrt{2} \right), \\ -{1 \over 2}\sqrt{2} & = 2 + {1 \over 2}\sqrt{2} + c \\ - 2 - \sqrt{2} & = c \\ \\ \therefore y & = x -2 - \sqrt{2} \\ \\ \\ \text{Using } & \left(2 - {1 \over 2}\sqrt{2}, {1 \over 2}\sqrt{2} \right), \\ {1 \over 2}\sqrt{2} & = 2 - {1 \over 2}\sqrt{2} + c \\ - 2 + \sqrt{2} & = c \\ \\ \therefore y & = x - 2 + \sqrt{2} \end{align}

(i)

\begin{align} x^2 + y^2 - 6x - 2y - 15 & = 0 \\ x^2 + y^2 + 2(-3)(x) + (2)(-1)(y) - 15 & = 0 \\ \\ \text{Comparing with } x^2 + y^2 & + 2gx + 2fy + c = 0, \\ g & = -3 \\ f & = -1 \\ c & = - 15 \\ \\ \therefore \text{ Centre } & (3, 1) \end{align}

(ii) The perpendicular bisector of chord AB passes through the centre of the circle (Circle property from E Maths)

\begin{align} \text{Let centre } & (3, 1) \text{ be } F. \\ \\ \text{Gradient of } MF & = {3 - 1 \over 2 - 3} \\ & = -2 \\ \\ \text{Gradient of } AB \times -2 & = -1 \\ \text{Gradient of } AB & = {-1 \over -2} \\ & = {1 \over 2} \\ \\ y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Using } & M(2, 3), \\ 3 & = {1 \over 2}(2) + c \\ 3 & = 1 + c \\ 2 & = c \\ \\ \text{Eqn of } AB: & \phantom{(} y = {1 \over 2}x + 2 \\ & 2y = x + 4 \end{align}

(iii) Since CD is parallel to AB, both lines have the same gradient. In addition, since CD is the diameter of the circle, it passes through the centre of the circle (3, 1).

\begin{align} y & = mx + c \\ y & = {1 \over 2}x + c \\ \\ \text{Using } & \text{centre (3, 1),} \\ 1 & = {1 \over 2}(3) + c \\ 1 & = {3 \over 2} + c \\ -{1 \over 2} & = c \\ \\ \text{Eqn of } CD: \phantom{0} y & = {1 \over 2}x - {1 \over 2} \phantom{000} \text{ --- (1)} \\ \\ \text{Eqn of circle: } \phantom{0} x^2 + y^2 & - 6x - 2y - 15 = 0 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + \left({1 \over 2}x - {1 \over 2}\right)^2 - 6x - 2\left({1 \over 2}x - {1 \over 2}\right) - 15 & = 0 \\ x^2 + \left({1 \over 2}x\right)^2 - 2\left({1 \over 2}x\right)\left({1 \over 2}\right) + \left(1 \over 2\right)^2 - 6x - x + 1 - 15 & = 0 \\ x^2 + {1 \over 4}x^2 - {1 \over 2}x + {1 \over 4} - 7x - 14 & = 0 \\ {5 \over 4}x^2 - {15 \over 2}x - {55 \over 4} & = 0 \\ 5x^2 - 30x - 55 & = 0 \\ x^2 - 6x - 11 & = 0 \\ \\ x & = {-b \pm \sqrt{ b^2 - 4ac} \over 2a} \\ & = {- (-6) \pm \sqrt{ (-6)^2 - 4(1)(-11) } \over 2(1) } \\ & = {6 \pm \sqrt{ 80 } \over 2} \\ & = {6 \pm \sqrt{16}\sqrt{5} \over 2} \\ & = {6 \pm 4 \sqrt{5} \over 2} \\ & = 3 \pm 2\sqrt{5} \\ \\ \text{Substitute } & x = 3 + 2\sqrt{5} \text{ into (1),} \\ y & = {1 \over 2} (3 + 2\sqrt{5}) - {1 \over 2} \\ & = {3 \over 2} + \sqrt{5} - {1 \over 2} \\ & = 1 + \sqrt{5} \\ \\ \therefore & \phantom{0} (3 + 2\sqrt{5}, 1 + \sqrt{5}) \\ \\ \text{Substitute } & x = 3 - 2\sqrt{5} \text{ into (1),} \\ y & = {1 \over 2} (3 - 2\sqrt{5}) - {1 \over 2} \\ & = {3 \over 2} - \sqrt{5} - {1 \over 2} \\ & = 1 - \sqrt{5} \\ \\ \therefore & \phantom{0} (3 - 2\sqrt{5}, 1 - \sqrt{5}) \end{align}

(i)

\begin{align} x^2 + y^2 - 4x - 2y + 1 & = 0 \phantom{000} \text{ --- (1)} \\ \\ kx - y & = k + 1 \\ - y & = - kx + k + 1 \\ y & = kx - k - 1 \phantom{000} \text{ --- (2)} \end{align} \begin{align} \text{Substitute } & \text{(2) into (1),} \\ x^2 + (kx - k - 1)^2 - 4x - 2(kx - k - 1) + 1 & = 0 \\ x^2 + (kx - k - 1)(kx - k - 1) - 4x - 2kx + 2k + 2 + 1 & = 0 \\ x^2 + k^2x^2 - k^2x - kx - k^2x + k^2 + k - kx + k + 1 - 4x - 2kx + 2k + 3 & = 0 \\ x^2 + k^2x^2 - 2k^2x - 4kx - 4x + k^2 + 4k + 4 & = 0 \\ (1 + k^2)x^2 + (-2k^2 -4k - 4)x + (k^2 + 4k + 4) & = 0 \end{align} \begin{align} b^2 - 4ac & = (-2k^2 - 4k - 4)^2 - 4(1 + k^2)(k^2 + 4k + 4) \\ & = (-2k^2 - 4k - 4)(-2k^2 - 4k - 4) - 4(k^2 + 4k + 4 + k^4 + 4k^3 + 4k^2) \\ & = 4k^4 + 8k^3 + 8k^2 + 8k^3 + 16k^2 + 16k + 8k^2 + 16k + 16 - 4(k^4 + 4k^3 + 5k^2 + 4k + 4) \\ & = 4k^4 + 16k^3 + 32k^2 + 32k + 16 - 4k^4 - 16k^3 - 20k^2 - 16k - 16 \\ & = 12k^2 + 16k \\ & = 4k(3k + 4) \\ \\ b^2 - 4ac & > 0 \phantom{00000} [\text{2 distinct points}] \\ 4k(3k + 4) & > 0 \\ k(3k + 4) & > 0 \end{align}

$$ k < -{4 \over 3} \phantom{0} \text{ or } \phantom{0} k > 0 $$

(ii)

\begin{align} b^2 - 4ac & = 0 \phantom{00000} [\text{1 point}] \\ 4k(3k + 4) & = 0 \\ k(3k + 4) & = 0 \\ \\ k = 0 \phantom{00}&\text{or}\phantom{00} 3k + 4 = 0 \\ & \phantom{or00+4} 3k = - 4 \\ & \phantom{or00+43} k = -{4 \over 3} \end{align}

(iii)

\begin{align} b^2 - 4ac & < 0 \phantom{00000} [\text{no points of intersection}] \\ 4k(3k + 4) & < 0 \\ k(3k + 4) & < 0 \end{align}

$$ -{4 \over 3} < k < 0 $$