Coordinate Geometry Part3

From StpmWiki

Jump to: navigation, search

Contents

Exercise 9

1) Find coordinates of point(s) on the given curves with the given condition

a) x=2t-1,y=3t\,

  • i) t=0\,
  • \left(-1,0\right)
    • t=0, x=-1, y=0, \therefore \left(-1,0\right)
  • ii) t=2\,
  • \left(3,6\right)
    • t=2, x=3, y=6, \therefore \left(3,6\right)

b) P\left(t^{2}-1,\frac{3}{t+1}\right)

  • i) t=p\,
  • \left(p^{2}-1,\frac{3}{p+1}\right)
    • t=p, x=p^{2}-1, y=\frac{3}{p+1}, \therefore \left(p^{2}-1,\frac{3}{p+1}\right)
  • ii) t=p-1\,
  • \left(p^{2}-2p,\frac{3}{p} \right)
    • \begin{align} & t=p-1 \\ & x=\left(p-1\right)^{2}-1=p^{2}-2p\\ & y =\frac{3}{\left(p-1\right)+1}=\frac{3}{p} \\ & \therefore \left(p^{2}-2p,\frac{3}{p} \right) \end{align}
  • iii) t=2p\,
  • \left(4p^{2}-1,\frac{3}{2p+1} \right)
    • \begin{align} & t=2p \\ & x=\left(2p\right)^{2}-1=4p^{2}-1\\ & y =\frac{3}{\left(2p\right)+1}=\frac{3}{2p+1} \\ & \therefore \left(4p^{2}-1,\frac{3}{2p+1} \right) \end{align}

c) x=3t-2,y=t^{2}\,

  • i) x=0\,
  • \left(0,\frac{4}{9}\right)
    • \begin{align} & x=0, \therefore 3t-2=0, t=\frac{2}{3} \\ & y = \frac{4}{9}, \therefore \left(0,\frac{4}{9}\right) \end{align}
  • ii) y=0\,
  • \left(-2,0\right)
    • \begin{align} & y =0, \therefore t^{2}=0, t=0 \\ & x = -2, \therefore \left(-2,0\right) \end{align}
  • iii) y=1\,
  • \left(1,1\right), \left(-5,1\right)
    • \begin{align} & y =1, \therefore t^{2}=1, t=\pm 1 \\ & \mbox{When } t= 1, x=1 ; t=-1, x=-5 \\ & \therefore \left(1,1\right), \left(-5,1\right) \end{align}

d) x=t^{2}-2t,y=\frac{2}{t^{2}}

  • i) x=-1\,
  • \left(-1,2\right)
    • \begin{align} & x=-1, \therefore t^{2}-2t=-1 \\ & t^{2}-2t+1=0 \\ & \left(t-1\right)^{2} =0 \\ & t=1,\therefore y = 2, \therefore \left(-1,2\right) \end{align}
  • ii) y=8\,
  • \left(-\frac{3}{4},8\right),\left(\frac{5}{4},8\right)
    • \begin{align} & y=8, \therefore \frac{2}{t^{2}}=8 \\ & t^{2}=\frac{1}{4}, t=\pm\frac{1}{2} \\ & \mbox{When } t=\frac{1}{2}, x=\frac{1}{4}-1=-\frac{3}{4} \\ & \mbox{When } t=-\frac{1}{2}, x=\frac{1}{4}+1=\frac{5}{4} \\ & \therefore \left(-\frac{3}{4},8\right),\left(\frac{5}{4},8\right)\\ \end{align}


2) Find Cartesian equation of the following parametric equations

a) x=1-t,y=3t+2\,

  • y=-3x+5 \,
    • \begin{align} & x=1-t\qquad\qquad y=3t+2\frac{\quad}{}(2) \\ & t=1-x\frac{\quad}{}(1)\\ & (1)\to(2) : y=3\left(1-x\right)+2\\ & y=-3x+5 \\ \end{align}

b) x=\frac{2}{t^{2}-1},y=t-2\,

  • x=\frac{2}{\left(y+3\right)\left(y+1\right)} \mbox{ or } \frac{2}{y^{2}+4y+3}
    • \begin{align} & x=\frac{2}{t^{2}-1}\frac{\quad}{}(1)\qquad\qquad y=t-2 \\ & \qquad \qquad \qquad \qquad \qquad \quad t=y+2 \frac{\quad}{}(2)\\ & (2)\to(1) :\\ & x=\frac{2}{\left(y+2\right)^{2}-1} \\ & x=\frac{2}{y^{2}+4y+3}\\ & x=\frac{2}{\left(y+3\right)\left(y+1\right)}\\ \end{align}

c) x=3-\frac{2}{t^{2}},y=\frac{1}{1-t}\,

  • x=\frac{y^{2}-6y+3}{\left(y-1\right)^{2}} \mbox{ or }x=3-\frac{2y^{2}}{\left(y-1\right)^{2}}
    • \begin{align} & x=3-\frac{2}{t^{2}}\frac{\quad}{}(1)\qquad\qquad y=\frac{1}{1-t} \\ & \qquad \qquad \qquad \qquad \qquad \quad 1-t=\frac{1}{y}\\ & \qquad \qquad \qquad \qquad \qquad \quad t=1-\frac{1}{y}\\ & \qquad \qquad \qquad \qquad \qquad \quad t=\frac{y-1}{y}\frac{\quad}{}(2)\\ & (2)\to(1) :\\ & x=3-\frac{2}{\left(\frac{y-1}{y}\right)^{2}} \\ & x=3-\frac{2y^{2}}{\left(y-1\right)^{2}}\\ & x=\frac{3y^{2}-6y+3-2y^{2}}{\left(y-1\right)^{2}}\\ & x=\frac{y^{2}-6y+3}{\left(y-1\right)^{2}}\\ \end{align}

d) x=1-t^{2},y=\frac{2}{3t^{2}}\,

  • y=\frac{2}{3\left(1-x\right)}\,
    • \begin{align} & x=1-t^{2}\qquad\qquad y=\frac{2}{3t^{2}}\frac{\quad}{}(2) \\ & t^{2}=1-x\frac{\quad}{}(1)\\ & (2)\to(1) : y=\frac{2}{3\left(1-x\right)}\\ \end{align}

e) x=\frac{2t-3}{t+2},y=\frac{2t}{t+3}\,

  • y=\frac{9y-6}{y+4}\,
    • \begin{align} & x=\frac{2t-3}{t+2}\frac{\quad}{}(1)\qquad y=\frac{2t}{t+3} \\ & \qquad \qquad \qquad \qquad \quad y\left(t+3\right)=2t\\ & \qquad \qquad \qquad \qquad \quad ty+3y=2t\\ & \qquad \qquad \qquad \qquad \quad 2t-ty=3y\\ & \qquad \qquad \qquad \qquad \quad t\left(2-y\right)=3y\\ & \qquad \qquad \qquad \qquad \quad t=\frac{3y}{2-y}\frac{\quad}{}(2)\\ & (2)\to(1) : y=\frac{2\left(\frac{3y}{2-y}\right)-3}{\left(\frac{3y}{2-y}\right)+2}\\ & y=\frac{\left[\frac{6y-3\left(2-y\right)}{2-y}\right]}{\left[\frac{3y+2\left(2-y\right)}{2-y}\right]}\\ & \quad =\frac{9y-6}{y+4}\\ \end{align}

f) x=t^{2}+3, y=t\left(t+1\right)

  • x^{2}+y^{2}-2xy-7x+6y+12=0\,
    • \begin{align} & x=t^{2}+3\frac{\quad}{}(1)\qquad y=t\left(t+1\right)\\ & \qquad \qquad \qquad \qquad \quad y=t^{2}+t\frac{\quad}{}(2)\\ & (2)\to(1) : y-x=t-3\\ & t=y-x+3\frac{\quad}{}(3)\\ & (3)\to(1) : x=\left(y-x+3\right)^{2}+3 \\ & x=\left(y-x+3\right)\left(y-x+3\right)+3 \\ & x =y^{2}+x^{2}-2xy+6y-6x+9+3 \\ & x^{2}+y^{2}-2xy-7x+6y+12=0 \\ \end{align}

g) x=\frac{2}{t^{2}},y=\frac{t-1}{t}

  • x=2\left(1-y\right)^{2}\mbox{ or }x=2\left(y-1\right)^{2}\,
    • \begin{align} & x=\frac{2}{t^{2}}\frac{\quad}{}(1)\qquad y=\frac{t-1}{t} \\ & \qquad \qquad \qquad \qquad \quad ty=t-1\\ & \qquad \qquad \qquad \qquad \quad t-ty=1\\ & \qquad \qquad \qquad \qquad \quad t\left(1-y\right)=1\\ & \qquad \qquad \qquad \qquad \quad t=\frac{1}{1-y}\frac{\quad}{}(2)\\ & (2)\to(1) : x=\frac{2}{\left(\frac{1}{1-y}\right)^{2}}\\ & \qquad x=2\left(1-y\right)^{2}\\ \end{align}

h) x=3t^{2}, y=4t^{3}-1\,

  • 16x^{3} =27\left(y+1\right)^{2}
    • \begin{align} & x=3t^{2}\qquad \qquad \quad y=4t^{3}-1 \\ & t^{2}=\frac{x}{3}\qquad \qquad \quad t^{3}=\frac{y+1}{4} \\ & t^{6}=\frac{x^{3}}{27}\frac{\quad}{}(1) \qquad t^{6}=\frac{\left(y+1\right)^{2}}{16} \frac{\quad}{}(2)\\ & (1)\to(2): \frac{x^{3}}{27} =\frac{\left(y+1\right)^{2}}{16}\\ & 16x^{3} =27\left(y+1\right)^{2} \end{align}

i) x=\frac{t}{t^{2}+1}, y=\frac{2}{t^{2}+1}

  • 4x^{2}+y^{2}-2y=0\,
    • \begin{align} & x=\frac{t}{t^{2}+1}\frac{\quad}{}(1) \qquad y=\frac{2}{t^{2}+1} \frac{\quad}{}(2)\\ & \frac{(1)}{(2)}: \frac{x}{y} =\frac{t}{2}\\ &\therefore t = \frac{2x}{y}\frac{\quad}{}(3)\\ & (3) \to (2) : y=\frac{2}{\left(\frac{2x}{y}\right)^{2}+1}\\ & y=\frac{2}{\left(\frac{4x^{2}+y^{2}}{y^{2}}\right)}\\ & y=\frac{2y^{2}}{4x^{2}+y^{2}}\\ & 1=\frac{2y}{4x^{2}+y^{2}}\\ & 4x^{2}+y^{2}=2y\\ & 4x^{2}+y^{2}-2y=0\\ \end{align}

j) x=\frac{t}{t+1}, y=\frac{2}{t+1}

  • 2x+y-2=0\,
    • \begin{align} & x=\frac{t}{t+1}\frac{\quad}{}(1) \qquad y=\frac{2}{t+1}\\ & \qquad \qquad \qquad \qquad \quad t+1=\frac{2}{y} \\ & \qquad \qquad \qquad \qquad \quad t=\frac{2}{y}-1 \\ & \qquad \qquad \qquad \qquad \quad t=\frac{2-y}{y} \frac{\quad}{}(2) \\ & (2) \to (1) : x=\frac{\left(\frac{2-y}{y}\right)}{\left(\frac{2-y}{y}\right)+1} \\ & x=\frac{\left(\frac{2-y}{y}\right)}{\left(\frac{2}{y}\right)} \\ & x =\frac{2-y}{2} \\ & 2x = 2-y \\ & y+2x-2=0 \\ \end{align}
    • OR
    • \begin{align} & x=\frac{t}{t+1}\frac{\quad}{}(1) \qquad y=\frac{2}{t+1} \frac{\quad}{}(2)\\ & \frac{(1)}{(2)}: \frac{x}{y} =\frac{t}{2}\\ &\therefore t = \frac{2x}{y}\frac{\quad}{}(3)\\ & (3) \to (2) : y=\frac{2}{\frac{2x}{y}+1}\\ & y=\frac{2}{\left(\frac{2x+y}{y}\right)}\\ & y=\frac{2y}{2x+y}\\ & 1=\frac{2}{2x+y}\\ & 2x+y-2=0\\ \end{align}


3) Show that point P\left(\frac{2t}{t^{2}+1},\frac{2t}{t^{2}-1}\right) lies on the curve y^{2}-x^{2}=x^{2}y^{2}\,

  • 4x^{2}+y^{2}-2y=0\,
    • \begin{align} & x=\frac{2t}{t^{2}+1}\qquad \qquad \qquad \qquad y=\frac{2t}{t^{2}-1}\\ & x^{2}=\left(\frac{2t}{t^{2}+1}\right)^{2}\qquad \qquad \quad \; y^{2}=\left(\frac{2t}{t^{2}-1}\right)^{2}\\ & x^{2}=\frac{4t^{2}}{t^{4}+2t^{2}+1}\qquad \qquad \quad y^{2}=\frac{4t^{2}}{t^{4}-2t^{2}+1}\\ & \frac{1}{x^{2}}=\frac{t^{4}+2t^{2}+1}{4t^{2}}\frac{\quad}{}(1) \qquad\; \frac{1}{y^{2}}=\frac{t^{4}-2t^{2}+1}{4t^{2}}\frac{\quad}{}(2)\\ & (1) - (2) : \frac{1}{x^{2}}-\frac{1}{y^{2}}=\frac{\left(t^{4}+2t^{2}+1\right)-\left(t^{4}-2t^{2}+1\right)}{4t^{2}}\\ & \qquad \qquad \quad \frac{1}{x^{2}}-\frac{1}{y^{2}}=\frac{4t^{2}}{4t^{2}}\\ & \qquad \qquad \quad \frac{1}{x^{2}}-\frac{1}{y^{2}}=1\\ & \qquad \qquad \quad \frac{y^{2}-x^{2}}{x^{2}y^{2}}=1\\ & \qquad \qquad \quad y^{2}-x^{2}=x^{2}y^{2}\frac{\qquad}{}\mbox{Shown} \end{align}

4) Verify that the Cartesian equation for the locus P\, that moves such that x=at^{2},y=2at\, is y^{2}-4ax=0\,

    • \begin{align} & y^{2}-4ax=0 \\ & \mbox{Substitute } x=at^{2}, y=2at \\ & \mbox{L.H.S. }=\left(2at\right)^{2}-4a\left(at^{2}\right)=4a^{2}t^{2}-4a^{2}t^{2}=0=\mbox{R.H.S }\\ & \therefore \mbox{the Cartesian equation is } y^{2}-4ax=0 \frac{\qquad}{}\mbox{(verified)} \end{align}


Exercise 10

1) Find coordinates of point of intersection of the given curves.

a) 4x+3y+15=0;x=2t-3,y=3t+1\,

  • \left(-\frac{63}{17},-\frac{1}{17}\right)
    • \begin{align} & 4x+3y+15=0\frac{\quad}{}(1)\qquad x=2t-3,y=3t+1 \frac{\quad}{}(2)\\ & (2) \to (1) :\\ & 4\left(2t-3\right)+3\left(3t+1\right)+15=0 \\ & 8t-12+9t+3+15 =0 \\ & 17t+6=0 \\ & t=-\frac{6}{17}\\ & x=2\left(-\frac{6}{17}\right)-3=-\frac{63}{17}\\ & y=3\left(-\frac{6}{17}\right)+1=-\frac{1}{17}\\ & \therefore \mbox{Point of intersection is } \left(-\frac{63}{17},-\frac{1}{17}\right) \end{align}
    • Check : 4\left(-\frac{63}{17}\right)+3\left(-\frac{1}{17}\right)+15=0

b) 3x+2y=0;x=t^{2}+t,y=t-1\,

  • \left(\frac{4}{9},-\frac{2}{3}\right), \left(2,-3\right)
    • \begin{align} & 3x+2y=0\frac{\quad}{}(1)\qquad x=t^{2}+t,y=t-1 \frac{\quad}{}(2)\\ & (2) \to (1) :\\ & 3\left(t^{2}+t\right)+2\left(t-1\right)=0 \\ & 3t^{2}+3t+2t-2 =0 \\ & 3t^{2}+5t-2=0 \\ & \left(3t-1\right)\left(t+2\right)=0 \\ & t=\frac{1}{3}, t=-2\\ & \mbox{When } t=\frac{1}{3} \\ & x=\left(\frac{1}{3}\right)^{2}+\frac{1}{3}=\frac{4}{9}\\ & y=\frac{1}{3}-1=-\frac{2}{3}\\ & \mbox{When } t=-2 \\ & x=\left(-2\right)^{2}-2=2\\ & y=-2-1=3\\ & \therefore \mbox{Point of intersections are } \left(\frac{4}{9},-\frac{2}{3}\right), \left(2,-3\right)\\ \end{align}
    • Check : 3\left(\frac{4}{9}\right)+2\left(-\frac{2}{3}\right)=0, 3\left(2\right)+2\left(-3\right)=0

c) x=\frac{6}{t}, y=t-2;x+y-3=0\,

  • \left(2,1\right), \left(3,0\right)
    • \begin{align} & x=\frac{6}{t}, y=t-2\frac{\quad}{}(1)\qquad x+y-3=0 \frac{\quad}{}(2)\\ & (1) \to (2) :\\ & \frac{6}{t}+t-2-3=0 \\ & 6+t^{2}-5t=0 \\ & t^{2}-5t+6=0 \\ & \left(t-3\right)\left(t-2\right)=0 \\ & t=3, t=2\\ & \mbox{When } t=3, x=2, y=1 \\ & \mbox{When } t=2, x=3, y=2 \\ & \therefore \mbox{Point of intersections are } \left(2,1\right), \left(3,0\right)\\ \end{align}
    • Check : 2+1-3=0, 3+0-3=0\,

d) x=\frac{1}{t^{2}}, y=\frac{t+1}{t};4x-3y+2=0\,

  • \left(\frac{1}{16},\frac{3}{4}\right), \left(1,2\right)
    • \begin{align} & x=\frac{1}{t^{2}}, y=\frac{t+1}{t}\frac{\quad}{}(1)\qquad 4x-3y+2=0 \frac{\quad}{}(2)\\ & (1) \to (2) :\\ & \frac{4}{t^{2}}-3\left(\frac{t+1}{t}\right)+2=0 \\ & 4-3t\left(t+1\right)+2t^{2}=0 \\ & -t^{2}-3t+4=0\\ & t^{2}+3t-4=0 \\ & \left(t+4\right)\left(t-1\right)=0 \\ & t=-4, t=1\\ & \mbox{When } t=4, x=\frac{1}{16}, y=\frac{3}{4} \\ & \mbox{When } t=1, x=1, y=2 \\ & \therefore \mbox{Point of intersections are } \left(\frac{1}{16},\frac{3}{4}\right), \left(1,2\right)\\ \end{align}
    • Check : 4\left(\frac{1}{16}\right)-3\left(\frac{3}{4}\right)+2=0, 4-6+2=0

e) xy=-5; x=t-7, y=t-1\,

  • \left(-1,5\right), \left(-5,1\right)
    • \begin{align} & xy=-5\frac{\quad}{}(1)\qquad x=t-7,y=t-1 \frac{\quad}{}(2)\\ & (2) \to (1) :\\ & \left(t-7\right)\left(t-1\right)=-5 \\ & t^{2}-8t+7 =-5 \\ & t^{2}-8t+12 =0 \\ & \left(t-6\right)\left(t-2\right)=0 \\ & t=6, t=2\\ & \mbox{When } t=6, x=-1, y=5 \\ & \mbox{When } t=2, x=-5, y=1 \\ & \therefore \mbox{Point of intersections are } \left(-1,5\right), \left(-5,1\right)\\ \end{align}
    • Check : \left(-1\right)\left(5\right)=-5, \left(-5\right)\left(1\right)=-5

f) x=t^{3}-{t}, y=t^{2}-1;x+2y=0\,

  • \left(0,0 \right), \left(-6,3\right)
    • \begin{align} & x=t^{3}-{t}, y=t^{2}-1\frac{\quad}{}(1)\qquad x+2y=0 \frac{\quad}{}(2)\\ & (1) \to (2) :\\ & t^{3}-{t}+2\left(t^{2}-1\right)=0 \\ & t^{3}+2t^{2}-t-2=0 \\ & \left(t-1\right)\left(t^{2}+3t+2\right)=0\\ & \left(t-1\right)\left(t+1\right)\left(t+2\right)=0\\ & t=1, t=-1, t=-2 \\ & \mbox{When } t=1, x=0, y=0 \\ & \mbox{When } t=-1, x=0, y=0 \\ & \mbox{When } t=-2, x=-6, y=3 \\ & \therefore \mbox{Point of intersections are } \left(0,0 \right), \left(-6,3\right)\\ \end{align}
    • Check : 0+0=0, -6+2\left(3\right)=0


2) Find the locus of P\, in each of the following case

a) A line with gradient m\, passes through the point \left(3,-1\right). This line intersects the x\,-axis and y\,-axis at A\, and B\, respectively. P\, divides AB\, in the ratio 1:2\,.

  • x = \frac{6y}{3y+1}
    • \begin{align} & l: m, \left(3,-1\right) \\ & y+1=m\left(x-3\right) \\ & y=mx-3m-1 \\ & \mbox{at } A, y=0, x=\frac{3m+1}{m}, A\left(\frac{3m+1}{m}, 0 \right)\\ & \mbox{at } B, x=0, y=-3m-1, B\left(0, -3m-1 \right)\\ & P =\left(\frac{0+2\left(\frac{3m+1}{m}\right)}{1+2},\frac{\left(-3m-1\right)+0}{1+2}\right)\\ & \quad =\left(\frac{2\left(3m+1\right)}{3m},\frac{-3m-1}{3}\right)\\ & x= \frac{2\left(3m+1\right)}{3m}\frac{\quad}{}(1) \qquad y =\frac{-3m-1}{3}\\ & \qquad \qquad \qquad \qquad \qquad \quad 3y=-3m-1 \\ & \qquad \qquad \qquad \qquad \qquad \quad 3m=-3y-1 \\ & \qquad \qquad \qquad \qquad \qquad \quad m=\frac{-3y-1}{3}\frac{\quad}{}(2) \\ & (2) \to (1) :x= \frac{2\left[3\left(\frac{-3y-1}{3}\right)+1\right]}{3\left(\frac{-3y-1}{3}\right)}\\ & \qquad \qquad \quad x= \frac{2\left(-3y\right)}{-3y-1}\\ & \qquad \qquad \quad x = \frac{6y}{3y+1} \end{align}

b) Given that T\left(t,0\right) and R\left(2,1\right). A line passes through T\, and perpendicular to TR\, meets y\,-axis at S\,. S\, is midpoint of TP\,.

  • y=-2x\left(x+2\right)
    • \begin{align} & m_{TR} =\frac{1}{2-t} \\ & l_{1} \perp TR \\ & m_{1} = t-2 \\ & l_{1}: m_{1} = t-2, T\left(t,0\right): \\ & y-0=\left(t-2\right)\left(x-t\right) \\ & y= tx-t^{2}-2x+2t \\ & y= tx-2x+2t-t^{2} \\ & \mbox{At }S, x=0, y=2t-t^{2} \\ & \therefore S\left(0,2t-t^{2}\right)\\ & \mbox{Let } P\left(x,y\right) \\ & S = \mbox{ Midpoint } TP \\ & \left(0,2t-t^{2}\right) = \left(\frac{x+t}{2},\frac{y}{2}\right)\\ & 0=\frac{x+t}{2} \qquad\qquad\qquad\; 2t-t^{2}=\frac{y}{2} \\ & t=-x\frac{\qquad}{}(1) \qquad\qquad 2t\left(2-t\right)=y\frac{\qquad}{}(2) \\ & (1)\to (2) : y=-2x\left(2+x\right)\\ & \qquad \qquad \therefore y=-2x\left(x+2\right) \end{align}

c) A line passes through the point \left(m^{2},2m\right) and has gradient -\frac{1}{m}. This line meets the line at 2y+x=4+2m^{2}\, at P\,.

  • x=2\left(y-2\right)\left(y-3\right)
    • \begin{align} & l_{1} :-\frac{1}{m}, \left(m^{2},2m\right):\\ & y-2m=-\frac{1}{m}\left(x-m^{2}\right) \\ & y-2m = -\frac{1}{m}x+m \\ & y = -\frac{1}{m}x+3m \\ & \mbox{To find } P :\\ & y = -\frac{1}{m}x+3m \frac{\quad}{}(1) :\\ & 2y+x=4+2m^{2} \frac{\quad}{}(2) :\\ & (1)\to (2) : \\ & 2\left(-\frac{1}{m}x+3m \right)+x=4+2m^{2} \\ & -2x+6m^{2}+mx=4m+2m^{3} \\ & mx-2x=2m^{3}-6m^{2}+4m \\ & x\left(m-2\right)=2m\left(m^{2}-3m+2\right) \\ & x=\frac{2m\left(m-2\right)\left(m-1\right)}{m-2} \\ & x=2m\left(m-1\right) \\ & \therefore y=-\frac{1}{m}\left[2m\left(m-1\right)\right]+3m\\ & y =-2\left(m-1\right)+3m\\ & y =-2m+2+3m\\ & y =m+2 \\ & \therefore P\left(2m\left(m-1\right), m+2\right) \\ & x=2m\left(m-1\right)\frac{\quad}{}(1) \qquad y =m+2 \\ & \qquad\qquad\qquad\qquad\qquad\quad\; m=y-2 \frac{\quad}{}(2)\\ & (2)-(1): x=2\left(y-2\right)\left(y-3\right)\\ \end{align}


d) A variable point T\, lies on the curve y^{2}=4x\,. P\, is the midpoint of TS\,, where S\left(2,0\right).

  • y^{2}=2\left(x-1\right)
    • \begin{align} & y^{2}=4x \\ & \mbox{Let } y=t, \therefore x =\frac{t^{2}}{4} \\ & \therefore T\left(\frac{t^{2}}{4},t\right)\\ & \mbox{Let } P\left(x,y\right) \\ & P = \mbox{ Midpoint of } TS \\ &\left(x,y\right) = \left(\frac{\frac{t^{2}}{4}+2}{2}, \frac{t}{2}\right)\\ & \therefore x=\frac{t^{2}+8}{8}\frac{\qquad}{}(1) \qquad y=\frac{t}{2}\\ & \qquad\qquad\qquad\qquad\qquad\quad t=2y\frac{\qquad}{}(2)\\ & (2) \to (1) : x=\frac{\left(2y\right)^{2}+8}{8} \\ & x=\frac{4y^{2}+8}{8}\\ & x=\frac{y^{2}+2}{2}\\ & 2x=y^{2}+2\\ & y^{2}=2x-2\\ & y^{2}=2\left(x-1\right)\\ \end{align}

e) Given that T\left(t+1,t^{2}\right),Q\left(2t,3\right). OPTQ\, is a parallelogram, where O\,is origin.

  • y =x^{2}-2x+2\,
    • \begin{align} & \mbox{Let } P\left(x,y\right) \\ & \mbox{Midpoint of } OT = \mbox{Midpoint of }PQ \\ & \left(\frac{t+1}{2},\frac{t^{2}}{2}\right)= \left(\frac{x+2t}{2},\frac{y+3}{2}\right)\\ & t+1 = x+2t \qquad \qquad \qquad t^{2}=y+3\frac{\qquad}{}(2) \\ & t=1-x\frac{\qquad}{}(1)\\ & (1) \to (2) : \left(1-x\right)^{2}=y+3 \\ & 1-2x+x^{2}=y+3 \\ & y =x^{2}-2x+2 \\ \end{align}
Retrieved from "http://www.stpmwiki.com/index.php/Coordinate_Geometry_Part3"