(a)

\begin{align} y^2 & = 4x \\ \\ \text{When } & x = 0, \\ y^2 & = 4(0) \\ & = 0 \\ y & = \sqrt{0} \\ & = 0 \\ \\ \implies & x \text{ and } y \text{-intercept is } (0, 0) \end{align}

$$ \text{Line of symmetry: } y = 0 $$

(b)

\begin{align} y^2 & = 0.5x \\ \\ \text{When } & x = 0, \\ y^2 & = 0.5(0) \\ & = 0 \\ y & = \sqrt{0} \\ & = 0 \\ \\ \implies & x \text{ and } y \text{-intercept is } (0, 0) \end{align}

$$ \text{Line of symmetry: } y = 0 $$

(c)

\begin{align} y^2 & = -2x \\ \\ \text{When } & x = 0, \\ y^2 & = -2(0) \\ & = 0 \\ y & = \sqrt{0} \\ & = 0 \\ \\ \implies & x \text{ and } y \text{-intercept is } (0, 0) \end{align}

$$ \text{Line of symmetry: } y = 0 $$

(i)

\begin{align} y^2 & = 4ax \\ \\ \text{Using } & (2, 4), \\ (4)^2 & = 4a(2) \\ 16 & = 8a \\ {16 \over 8} & = a \\ 2 & = a \end{align}

(ii)

\begin{align} y^2 & = 4(2)x \\ y^2 & = 8x \\ \\ \text{When } & x = 0, \\ y^2 & = 8(0) \\ & = 0 \\ y & = \sqrt{0} \\ & = 0 \\ \\ \implies & x \text{ and } y \text{-intercept is } (0, 0) \end{align}

\begin{align} 2y & = x + 3 \\ 2y - 3 & = x \\ x & = 2y - 3 \phantom{000} \text{ --- (1)} \\ \\ y^2 & = 2x + 3 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ y^2 & = 2(2y - 3) + 3 \\ y^2 & = 4y - 6 + 3 \\ y^2 - 4y + 3 & = 0 \\ (y - 3)(y - 1) & = 0 \\ \\ y - 3 = 0 \phantom{000} &\text{or} \phantom{000} y - 1 = 0 \\ y = 3 \phantom{000} & \phantom{or00000-} y = 1 \\ \\ \text{Substitute } & y = 3 \text{ into (1),} \\ x & = 2(3) - 3 \\ & = 3 \\ \\ \therefore & \phantom{0} (3, 3) \\ \\ \text{Substitute } & y = 1 \text{ into (1),} \\ x & = 2(1) - 3 \\ & = -1 \\ \\ \therefore & \phantom{0} (-1, 1) \end{align}

(i)

\begin{align} y & = 2x + 4 \phantom{000} \text{ --- (1)} \\ \\ y & = x^2 - 4 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x + 4 & = x^2 - 4 \\ 0 & = x^2 - 2x - 8 \\ 0 & = (x - 4)(x + 2) \\ \\ x - 4 = 0 \phantom{000} &\text{or} \phantom{000} x + 2 = 0 \\ x = 4 \phantom{000} & \phantom{or00000-} x = - 2 \\ \\ \text{Substitute } x & = 4 \text{ into (1),} \\ y & = 2(4) + 4 \\ & = 12 \\ \\ \therefore & \phantom{0} (4, 12) \\ \\ \text{Substitute } x & = -2 \text{ into (2),} \\ y & = 2(-2) + 4 \\ & = 0 \\ \\ \therefore & \phantom{0} (-2, 0) \end{align}

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

\begin{align} \text{Length of } AB & = \sqrt{(4 - (-2))^2 + (12 - 0)^2} \\ & = \sqrt{180} \\ & = \sqrt{36} \sqrt{5} \\ & = 6\sqrt{5} \text{ units} \end{align}

(i)

\begin{align} y & = mx \phantom{000} \text{ --- (1)} \\ \\ y^2 & = 2x - 1 \phantom{000} \text{ --- (2)} \\ \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (mx)^2 & = 2x - 1 \\ m^2x^2 & = 2x - 1 \\ m^2x^2 - 2x + 1 & = 0 \\ \\ b^2 - 4ac & = 0 \phantom{00000} [\text{1 real root}] \\ (-2)^2 - 4(m^2)(1) & = 0 \\ 4 - 4m^2 & = 0 \\ 1 - m^2 & = 0 \\ (1 - m)(1 + m) & = 0 \\ \\ 1 - m = 0 \phantom{000} &\text{or} \phantom{000} 1 + m = 0 \\ -m = -1 \phantom{0/} & \phantom{or00000+} m = -1 \\ m = 1 \phantom{0/-} & \end{align}

(ii)

\begin{align} \text{When } & m = 1, \\ y & = x \phantom{000} \text{ --- (1)} \\ \\ y^2 & = 2x - 1 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (x)^2 & = 2x - 1 \\ x^2 - 2x + 1 & = 0 \\ (x - 1)^2 & = 0 \\ x & = 1 \\ \\ \text{Substitute } x & = 1 \text{ into (1),} \\ y & = 1 \\ \\ \therefore & \phantom{0} (1, 1) \end{align}
\begin{align} \text{When } & m = -1, \\ y & = -x \phantom{000} \text{ --- (1)} \\ \\ y^2 & = 2x - 1 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (-x)^2 & = 2x - 1 \\ x^2 - 2x + 1 & = 0 \\ (x - 1)^2 & = 0 \\ x & = 1 \\ \\ \text{Substitute } x & = 1 \text{ into (1),} \\ y & = -(1) \\ & = -1 \\ \\ \therefore & \phantom{0} (1, -1) \end{align}

Graph for the entire question

(i)

\begin{align} \text{When } & x = 0, \\ y^2 & = 0 + 4 \\ & = 4 \\ y & = \pm \sqrt{4} \\ & = \pm 2 \\ \\ \implies & y \text{-intercepts are } (0, 2) \text{ and } (0, -2) \\ \\ \\ \text{When } & y = 0, \\ (0)^2 & = x + 4 \\ -x & = 4 \\ x & = -4 \\ \\ \implies & x \text{-intercept is } (-4, 0) \end{align}

(ii)

\begin{align} 4x^2 - 13x & = - 5 \\ 4x^2 - 13x + 9 & = -5 + 9 \\ 4x^2 - 13x + 9 & = 4 \\ 4x^2 - 13x + x + 9 & = 4 + x \\ 4x^2 - 12x + 9 & = x + 4 \\ (2x)^2 - 2(2x)(3) + 3^2 & = x + 4 \\ (2x - 3)^2 & = x + 4 \end{align}

(iii)

\begin{align} 4x^2 - 13x & = -5 \\ \\ \text{From } & \text{part ii,} \\ (2x - 3)^2 & = x + 4 \\ \\ \text{Since equation of parabola} & \text{ is } y^2 = x + 4, \\ (2x - 3)^2 & = y^2 \\ 2x - 3 & = y \\ \\ \therefore \text{Draw } y & = 2x - 3 \end{align}
\begin{align} \text{From the} & \text{ graph,} \\ \\ x = 0.45 \phantom{00} & \text{or} \phantom{00} x = 2.8 \end{align}

Graph for the entire question

(i)

\begin{align} \text{When } & x = 0, \\ y^2 & = 6(0) \\ & = 0 \\ y & = \sqrt{0} \\ & = 0 \\ \\ \implies & x \text{ and } y \text{-intercept is } (0,0) \end{align}

(ii)

\begin{align} \text{From } & \text{the graph,} \\ x & > 1.5 \end{align}

(iii)

\begin{align} 4x^2 - 10x + 1 & = 0 \\ 4x^2 - 10x + 1 + 6x & = 6x \\ 4x^2 - 4x + 1 & = 6x \\ (2x)^2 - 2(2x)(1) + (1)^2 & = 6x \\ (2x - 1)^2 & = 6x \\ \\ \text{Since equation of the} & \text{ parabola is } y^2 = 6x, \\ (2x - 1)^2 & = y^2 \\ 2x - 1 & = y \\ \\ \therefore \text{Draw } y & = 2x - 1 \end{align}
\begin{align} \text{From } & \text{the graph,} \\ x = 0.10 \phantom{00} & \text{or} \phantom{00} x = 2.4 \end{align}

(i)

\begin{align} y & = x + {2 \over x} \phantom{000} \text{ --- (1)} \\ \\ y & = 2x - 1 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x + {2 \over x} & = 2x - 1 \\ x\left(x + {2 \over x}\right) & = x(2x - 1) \\ x^2 + 2 & = 2x^2 - x \\ 0 & = x^2 - x - 2 \\ 0 & = (x - 2)(x + 1) \\ \\ x - 2 = 0 \phantom{00} &\text{or} \phantom{00} x + 1 = 0 \\ x = 2 \phantom{00} & \phantom{or00+1} x = -1 \\ \\ \text{Substitute } x & = 2 \text{ into (1),} \\ y & = (2) + {2 \over (2)} \\ & = 3 \\ \\ \therefore & \phantom{0} A(2, 3) \\ \\ \text{Substitute } x & = -1 \text{ into (1),} \\ y & = (-1) + {2 \over (-1)} \\ & = -3 \\ \\ \therefore & \phantom{0} B(-1, -3) \end{align}

(ii)

\begin{align} \text{Area of } \triangle PAB & = {1 \over 2} \left| \begin{matrix} 5 & 2 & -1 & 5 \\ 0 & 3 & -3 & 0 \end{matrix} \right| \\ & = {1 \over 2} [(5)(3)+(2)(-3)+(-1)(0)] - {1 \over 2}[(0)(2)+(3)(-1)+(-3)(5)] \\ & = 13{1 \over 2} \text{ sq. units} \end{align}

(i)

\begin{align} y & = {1 \over x} \phantom{000} \text{ --- (1)} \\ \\ y & = 2x + 1 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{ (1) into (2),} \\ {1 \over x} & = 2x + 1 \\ 1 & = x(2x + 1) \\ 1 & = 2x^2 + x \\ 0 & = 2x^2 + x - 1 \\ 0 & = (2x - 1)(x + 1) \\ \\ 2x - 1 = 0 \phantom{00} &\text{or} \phantom{00} x + 1 = 0 \\ 2x = 1 \phantom{00} &\text{or} \phantom{00+1} x = - 1 \\ x = {1 \over 2} \phantom{..} & \\ \\ \text{Substitute } & x = {1 \over 2} \text{ into (2),} \\ y & = 2\left(1 \over 2\right) +1 \\ & = 2 \\ \\ \therefore & \phantom{0} A\left({1 \over 2}, 2\right) \\ \\ \text{Substitute } & x = -1 \text{ into (2),} \\ y & = 2(-1) + 1 \\ & = -1 \\ \\ \therefore & \phantom{0} B(-1, -1) \end{align}

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

\begin{align} \text{When } & y = 0, \\ 0 & = 2x + 1 \\ -2x & = 1 \\ x & = {1 \over -2} \\ & = -{1 \over 2} \\ \\ \therefore & \phantom{0} M\left(-{1 \over 2}, 0\right) \\ \\ AM & = \sqrt{ \left[{1 \over 2} - \left(-{1 \over 2}\right)\right]^2 + (2 - 0)^2} \\ & = \sqrt{5} \text{ units} \\ \\ MB & = \sqrt{ \left[ -1 - \left(- {1 \over 2}\right) \right]^2 + (-1 - 0)^2 } \\ & = \sqrt{5 \over 4} \\ & = {\sqrt{5} \over \sqrt{4}} \\ & = {\sqrt{5} \over 2} \\ & = {1 \over 2} \sqrt{5} \text{ units} \\ \\ AM & : MB \\ \sqrt{5} & : {1 \over 2}\sqrt{5} \\ 1 & : {1 \over 2} \\ 2 & : 1 \end{align}

(i)

\begin{align} y^2 & = 3x \phantom{000} \text{ --- (1)} \\ \\ y & = 6 - x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (6 - x)^2 & = 3x \\ (6)^2 - 2(6)(x) + (x)^2 & = 3x \\ 36 - 12x + x^2 & = 3x \\ x^2 - 15x + 36 & = 0 \\ (x - 3)(x - 12) & = 0 \\ \\ x - 3 = 0 \phantom{00} &\text{or} \phantom{00} x - 12 = 0 \\ x = 3 \phantom{00} & \phantom{or00-12} x = 12 \\ \\ \text{Substitute } & x = 3 \text{ into (2),} \\ y & = 6 - (3) \\ & = 3 \\ \\ \therefore & \phantom{0} A(3, 3) \\ \\ \text{Substitute } & x = 12 \text{ into (2),} \\ y & = 6 - (12) \\ & = -6 \\ \\ \therefore & \phantom{0} B(12, -6) \end{align}

(ii)

\begin{align} \text{Mid-point of } AB, M & = \left( {3 + 12 \over 2}, {3 + (-6) \over 2} \right) \\ & = (7.5, -1.5) \\ \\ \text{Area of } \triangle OMB & = {1 \over 2} \left| \begin{matrix} 0 & 12 & 7.5 & 0 \\ 0 & -6 & -1.5 & 0 \end{matrix} \right| \\ & = {1 \over 2}[(0)(-6)+(12)(-1.5)+(7.5)(0)] \\ & \phantom{=.} - {1 \over 2}[(0)(12)+(-6)(7.5)+(-1.5)(0)] \\ & = 13{1 \over 2} \text{ sq. units} \end{align}

(i)

\begin{align} y^2 & = 12 - 2x \phantom{000} \text{ --- (1)} \\ \\ y & = 2 - x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (2 - x)^2 & = 12 - 2x \\ (2)^2 - 2(2)(x) + (x)^2 & = 12 - 2x \\ 4 - 4x + x^2 & = 12 - 2x \\ x^2 - 2x - 8 & = 0 \\ (x - 4)(x + 2) & = 0 \\ \\ x - 4 = 0 \phantom{00} &\text{or} \phantom{00} x + 2 = 0 \\ x = 4 \phantom{00} & \phantom{or00+2} x = - 2 \\ \\ \text{Substitute } & x = 4 \text{ into (2),} \\ y & = 2 - (4) \\ & = -2 \\ \\ \therefore & \phantom{0} B(4, -2) \\ \\ \text{Substitute } & x = -2 \text{ into (2),} \\ y & = 2 - (-2) \\ & = 4 \\ \\ \therefore & \phantom{0} A(-2, 4) \end{align}

(ii) Recall that the perpendicular bisector of AB passes through the midpoint of AB

\begin{align} \text{Mid-point of } AB & = \left( {4 + (-2) \over 2}, {(-2) + 4 \over 2} \right) \\ & = (1, 1) \\ \\ \text{Gradient of } AB & = {-2 - 4 \over 4 - (-2)} \\ & = -1 \\ \\ \text{Gradient of perp. bisector} \times -1 & = -1 \\ \text{Gradient of perp. bisector} & = 1 \\ \\ y & = mx + c \\ y & = x + c \\ \\ \text{Using } & (1, 1), \\ 1 & = 1 + c \\ 0 & = c \\ \\ \text{Eqn of perp. bisector of } & AB: \phantom{.} y = x \end{align}

(i)

\begin{align} y & = x^2 + x - 2 \phantom{000} \text{ --- (1)} \\ \\ y & = 2x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + x - 2 & = 2x \\ x^2 - x - 2 & = 0 \\ (x - 2)(x + 1) & = 0 \\ \\ x - 2 = 0 \phantom{00} &\text{or} \phantom{00} x + 1 = 0 \\ x = 2 \phantom{00} & \phantom{or00+1} x = - 1 \\ \\ \text{Substitute } & x = 2 \text{ into (2),} \\ y & = 2(2) \\ & = 4 \\ \\ \therefore & \phantom{0} A(2, 4) \\ \\ \text{Substitute } & x = -1 \text{ into (2),} \\ y & = 2(-1) \\ & = -2 \\ \\ \therefore & \phantom{0} B(-1, -2) \end{align}

(ii)

\begin{align} \text{Length of } AB & = \sqrt{ [2 - (-1)]^2 + [4 - (-2)]^2} \\ & = \sqrt{ 45 } \\ & = \sqrt{9} \sqrt{5} \\ & = 3\sqrt{5} \text{ units} \end{align}

(iii)

\begin{align} \text{Area of } \triangle ABC & = {1 \over 2} \left| \begin{matrix} 2 & -3 & -1 & 2 \\ 4 & 4 & -2 & 4 \end{matrix} \right| \\ & = {1 \over 2} [(2)(4)+(-3)(-2)+(-1)(4)] \\ & \phantom{=.} - {1 \over 2}[(4)(-3)+(4)(-1)+(-2)(2)] \\ & = 15 \text{ sq. units} \\ \\ \text{Area of } \triangle ABC & = {1 \over 2} \times AB \times h \\ 15 & = {1 \over 2} (3\sqrt{5})(h) \\ 30 & = 3\sqrt{5} h \\ 10 & = \sqrt{5} h \\ \\ h & = {10 \over \sqrt{5}} \\ & = {10 \over \sqrt{5}} \times {\sqrt{5} \over \sqrt{5}} \\ & = {10 \sqrt{5} \over 5} \\ & = 2\sqrt{5} \text{ units} \end{align}

(i)

\begin{align} y & = x + k \\ \\ \text{When } x = 2 & \text{ and } y = -4, \\ (-4) & = (2) + k \\ -4 - 2 & = k \\ -6 & = k \\ \\ y & = x + (-6) \\ y & = x - 6 \phantom{000} \text{ --- (1)} \\ \\ y^2 & = 8x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (x - 6)^2 & = 8x \\ (x)^2 - 2(x)(6) + (6)^2 & = 8x \\ x^2 - 12x + 36 & = 8x \\ x^2 - 20x + 36 & = 0 \\ (x - 2)(x - 18) & = 0 \\ \\ x - 2 = 0 \phantom{00}&\text{or} \phantom{00} x - 18 = 0 \\ x = 2 \phantom{00}& \phantom{or00-18} x = 18 \\ \\ \text{Substitute } & x = 18 \text{ into (1),} \\ y & = (18) - 6 \\ & = 12 \\ \\ \therefore & \phantom{0} B (18, 12) \end{align}

(ii) Recall that the perpendicular bisector of AB passes through the midpoint of AB

\begin{align} \text{Mid-point of } AB & = \left( {2 + 18 \over 2}, {-4 + 12 \over 2} \right) \\ & = (10, 4) \\ \\ \text{Gradient of } AB & = {-4 - 12 \over 2 - 18} \\ & = 1 \\ \\ \text{Gradient of perp. bisector} \times 1 & = -1 \\ \text{Gradient of perp. bisector} & = -1 \\ \\ y & = mx + c \\ y & = -x + c \\ \\ \text{Using } & (10, 4), \\ 4 & = -(10) + c \\ 4 & = -10 + c \\ 14 & = c \\ \\ \text{Eqn of perp. bisector of } & AB: \phantom{.} y = -x + 14 \\ & \phantom{0(} x + y = 14 \end{align}

(i)

\begin{align} y^2 & = 2x - 3 \phantom{000} \text{ --- (1)} \\ \\ 2x + 3y & = 7 \\ 2x & = 7 - 3y \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ y^2 & = 7 - 3y - 3 \\ y^2 & = 4 - 3y \\ y^2 + 3y - 4 & = 0 \\ (y + 4)(y - 1) & = 0 \\ \\ y + 4 = 0 \phantom{00}&\text{or}\phantom{00} y - 1 = 0 \\ y = - 4 \phantom{00}& \phantom{or00-1} y = 1 \\ \\ \text{Substitute } & y = -4 \text{ into (2),} \\ 2x & = 7 - 3(-4) \\ 2x & = 19 \\ x & = {19 \over 2} \\ \\ \therefore & \phantom{0} B \left({19 \over 2}, - 4\right) \\ \\ \text{Substitute } & y = 1 \text{ into (2),} \\ 2x & = 7 - 3(1) \\ 2x & = 4 \\ x & = {4 \over 2} \\ & = 2 \\ \\ \therefore & \phantom{0} A (2, 1) \\ \\ \text{Length of } AB & = \sqrt{ \left({19 \over 2} - 2\right)^2 + [1 - (-4)]^2 } \\ & = \sqrt{ 325 \over 4} \\ & = {\sqrt{325} \over \sqrt{4}} \\ & = {\sqrt{25} \sqrt{13} \over 2} \\ & = {5 \sqrt{13} \over 2} \text{ units (Shown)} \end{align}

(ii) To use 'shoelace' method, remember to take the points in an anti-clockwise order (i.e. P, A, B)

\begin{align} \text{Area of } \triangle PAB & = {1 \over 2} \left| \begin{matrix} 6 & 2 & {19 \over 2} & 6 \\ 3 & 1 & -4 & 3 \end{matrix} \right| \\ & = {1 \over 2} \left[(6)(1) + (2)(-4) + \left(19 \over 2\right)(3)\right] \\ & \phantom{=.} - {1 \over 2} \left[ (3)(2) + (1)\left(19 \over 2\right) + (-4)(6) \right] \\ & = 17{1 \over 2} \text{ sq. units} \end{align}

(iii)

\begin{align} \text{Area of } \triangle PAB & = {1 \over 2} \times AB \times h \\ 17{1 \over 2} & = {1 \over 2} \left(5\sqrt{13} \over 2\right)(h) \\ 35 & = \left(5\sqrt{13} \over 2\right)(h) \\ \\ h & = {35(2) \over 5\sqrt{13}} \\ & = {7(2) \over \sqrt{13}} \\ & = {14 \over \sqrt{13}} \\ \\ \therefore \text{Perp. distance} & = {14 \over \sqrt{13}} \text{ units} \end{align}

\begin{align} \text{Let coordinates} & \text{ of } B \text{ be } (a, b). \\ \\ \text{When } x = a & \text{ and } y = b, \\ (b)^2 & = 2(a) \\ b^2 & = 2a \\ \\ a & = {b^2 \over 2} \\ \\ \therefore & \phantom{0} B \left({b^2 \over 2}, b\right) \\ \\ \text{Mid-point of } AB & = \left( {2 + {b^2 \over 2} \over 2}, {6 + b \over 2} \right) \\ \\ x & = {2 + {b^2 \over 2} \over 2} \\ 2x & = 2 + {b^2 \over 2} \\ 4x & = 4 + b^2 \phantom{000} \text{ --- (1)} \\ \\ y & = {6 + b \over 2} \\ 2y & = 6 + b \\ \\ b & = 2y - 6 \phantom{000} \text{ --- (2)} \\ \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 4x & = 4 + (2y - 6)^2 \\ 4x & = 4 + (2y)^2 - 2(2y)(6) + (6)^2 \\ 4x & = 4 + 4y^2 - 24y + 36 \\ 0 & = 4y^2 - 24y - 4x + 40 \\ 0 & = y^2 - 6y - x + 10 \end{align}

(i)

\begin{align} y^2 & = 8x \\ \\ \text{When } x = a & \text{ and } y = b, \\ (b)^2 & = 8(a) \\ b^2 & = 8a \\ b & = \pm\sqrt{8a} \\ \\ b = \sqrt{8a} \phantom{00} &\text{or}\phantom{00} b = -\sqrt{8a} \text{ (Reject, as } P \text{ is above } x\text{-axis)} \\ \\ \therefore & \phantom{0} P(a, \sqrt{8a}) \\ \\ PF & = \sqrt{ (a - 2)^2 + (\sqrt{8a} - 0)^2 } \\ & = \sqrt{ (a)^2 - 2(a)(2) + (2)^2 + 8a } \\ & = \sqrt{ a^2 - 4a + 4 + 8a } \\ & = \sqrt{ a^2 + 4a + 4 } \\ & = \sqrt{ (a + 2)^2 } \\ & = (a + 2) \text{ units} \end{align}

(ii) Since PQ is a horizontal line, P and Q have the same y-coordinate

\begin{align} \text{Coordinates of } Q & = (-2, b) \\ & = (-2, \sqrt{8a}) \\ \\ PQ & = a - (-2) \\ & = (a + 2) \text{ units} \\ & = PF \text{ (Shown)} \end{align}

(iii) To use 'shoelace' method, remember to take the points in an anti-clockwise order (i.e. Q, F, P)

\begin{align} \text{Area of } \triangle PQF & = {1 \over 2} \left| \begin{matrix} -2 & 2 & a & -2 \\ \sqrt{8a} & 0 & \sqrt{8a} & \sqrt{8a} \end{matrix} \right| \\ & = {1 \over 2} [(-2)(0)+(2)(\sqrt{8a})+(a)(\sqrt{8a})] \\ & \phantom{=.} - {1 \over 2} [(\sqrt{8a})(2)+(0)(a)+(\sqrt{8a})(-2)] \\ & = {1 \over 2}(0 + 2\sqrt{8a} + a\sqrt{8a}) - {1 \over 2}(2\sqrt{8a} + 0 - 2\sqrt{8a}) \\ & = {1 \over 2}(2\sqrt{8a} + a\sqrt{8a}) \\ & = {1 \over 2}\sqrt{8a}(2 + a) \\ & = {1 \over 2} (\sqrt{4} \sqrt{2a}) (2 + a) \\ & = {1 \over 2}(2 \sqrt{2a})(2 + a) \\ & = \sqrt{2a}(a + 2) \phantom{0} \text{ sq. units} \end{align}

(i)

\begin{align} & \phantom{.} D(at^2, 2at) \\ \\ & \phantom{.} A(0, 2at) \\ \\ \\ x \text{-coordi} & \text{nate of } C = at^2 \\ \\ \text{Substitute } & x = at^2 \text{ into } y^2 = 4ax, \\ y^2 & = 4a(at^2) \\ y^2 & = 4a^2 t^2 \\ y & = \pm \sqrt{4a^2 t^2} \\ y & = \pm 2at \\ \\ \therefore & \phantom{.} C(at^2, -2at) \\ \\ & \phantom{.} B(0, - 2at) \end{align}

(ii)

\begin{align} \text{Gradient of } AC & = {2at - (-2at) \over 0 - at^2} \\ & = {4at \over - at^2} \\ & = -{4 \over t} \\ \\ y & = mx + c \\ y & = -{4 \over t}x + c \\ \\ \text{Using } & A(0, 2at), \\ c & = 2at \\ \\ \text{Eqn of } AC: & \phantom{.} y = -{4 \over t}x + 2at \phantom{00} \text{--- (1)} \\ \\ \text{Eqn of parabola: } & y^2 = 4ax \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ \left(-{4 \over t}x + 2at \right)^2 & = 4ax \\ \left(-{4 \over t}x\right)^2 + 2 \left(-{4 \over t}x\right)(2at) + (2at)^2 & = 4ax \\ {16 \over t^2} x^2 - {8 \over t}(x)(2at) + 4a^2 t^2 & = 4ax \\ {16 \over t^2} x^2 - 16ax + 4a^2 t^2 & = 4ax \\ {16 \over t^2} x^2 - 20ax + 4a^2 t^2 & = 0 \\ 16x^2 - 20at^2 x + 4a^2 t^4 & = 0 \\ 4x^2 - 5at^2 x + a^2 t^4 & = 0 \\ (4x - at^2)(x - at^2) & = 0 \end{align}\begin{align} 4x - at^2 & = 0 && \text{ or } x - at^2 & = 0 \\ 4x & = at^2 &&& x & = at^2 \phantom{0} \text{(Point } C) \\ x & = {at^2 \over 4} \end{align}
\begin{align} \text{Substitute } & x = {at^2 \over 4} \text{ into (1),} \\ y & = -{4 \over t} \left(at^2 \over 4\right) + 2at \\ y & = - at + 2at \\ y & = at \\ \\ \therefore \text{Point is } & \left( {at^2 \over 4}, at \right) \end{align}

(iii)

\begin{align} \text{By similar triangles, } AM & = {2 \over 3} AC \\ \\ \text{Horizontal distance between } A \text{ and } C & = at^2 \\ \\ \text{Horizontal distance between } A \text{ and } M & = {2 \over 3}at^2 \\ \\ x \text{-coordinate of } M & = 0 + {2 \over 3}at^2 \\ & = {2 \over 3}at^2 \\ \\ \\ \text{Vertical distance between } A \text{ and } C & = 2at - (-2at) \\ & = 4at \\ \\ \text{Vertical distance between } A \text{ and } M & = {2 \over 3}(4at) \\ & = {8 \over 3}at \\ \\ y \text{-coordinate of } M & = 2at - {8 \over 3}at \\ & = -{2 \over 3}at \\ \\ \therefore & \phantom{.} M \left({2 \over 3}at^2, -{2 \over 3}at\right) \end{align}

(iv)

\begin{align} \text{From (ii), gradient of } AC & = -{4 \over t} \\ \\ \text{Gradient of line} & = {-1 \over -{4 \over t}} \phantom{000000} [m_1 \times m_2 = -1] \\ & = {t \over 4} \\ \\ y & = mx + c \\ y & = {t \over 4}x + c \\ \\ \text{Using } & M \left({2 \over 3}at^2, -{2 \over 3}at\right), \\ -{2 \over 3}at & = {t \over 4} \left({2 \over 3}at^2\right) + c \\ -{2 \over 3}at & = {1 \over 6}at^3 + c \\ -{2 \over 3}at - {1 \over 6}at^3 & = c \\ \\ \therefore y & = {t \over 4}x - {2 \over 3}at - {1 \over 6}at^3 \end{align}

(i)

\begin{align} \text{Curve 1: } y^2 & = x \\ y & = \pm \sqrt{x} \\ \\ \text{Curve 2: } y & = \sqrt{x} \\ \\ \\ \text{Thus, graphs are} & \text{ not the same} \end{align}

(ii)

(i)

\begin{align} \text{Distance} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 } \\ & = \sqrt{ [-2.3 - (-0.1)]^2 + (1 - 0)^2 } \\ & = \sqrt{5.84} \text{ A.U.} \\ \\ \text{Distance in miles} & = \sqrt{5.84} \times (93 \phantom{.} 000 \phantom{.} 000) \\ & = 93 \phantom{.} 000 \phantom{.} 000 \sqrt{5.84} \\ \\ \text{Distance in km} & = 93 \phantom{.} 000 \phantom{.} 000 \sqrt{5.84} \times 1.609 \\ & \approx 3.61 \times 10^8 \text{ km} \end{align}

(ii)

\begin{align} \text{Let } (a, b) \text{ denote the } & \text{point where the comet is closest to the Sun} \\ \\ \text{Since comet's path} & \text{ is defined by } x = -2.3y^2, \\ a & = -2.3b^2 \\ \\ \text{Closest point } & \text{is } (-2.3b^2, b) \\ \\ \text{Distance, } d & = \sqrt{ [-2.3b^2 - (-0.1)]^2 + (b - 0)^2} \\ & = \sqrt{ (-2.3b^2 + 0.1)^2 + b^2 } \\ & = \sqrt{ (-2.3b^2)^2 + 2(-2.3b^2)(0.1) + (0.1)^2 + b^2 } \\ & = \sqrt{ 5.29b^4 - 0.46b^2 + 0.01 + b^2 } \\ & = \sqrt{ 5.29b^4 + 0.54b^2 + 0.01 } \\ \\ \text{Since } b^2 \ge 0 \text{ and } b^4 \ge 0, & \text{ shortest distance occurs when } b = 0 \\ \\ \text{Shortest distance} & = \sqrt{ 0 + 0 + 0.01} \\ & = 0.1 \text{ A.U.} \\ & = 0.1 \times 93 \phantom{.} 000 \phantom{.} 000 \times 1.609 \\ & = 14 \phantom{.} 963 \phantom{.} 700 \text{ km} \\ & = 14 \phantom{.} 963 \phantom{.} 700 \phantom{.} 000 \text{ m} \\ \\ V & = {k \over \sqrt{d}} \\ V & = { 1.17 \times 10^{10} \over \sqrt{14 \phantom{.} 963 \phantom{.} 700 \phantom{.} 000} } \\ V & \approx 95 \phantom{.} 600 \text{ m/s} \end{align}