mail  mail@stemandmusic.in
    
call  +91-9818088802
Donate

Converting Parabola Equation from Implicit Coordinate to General Parametric

  1. The Implicit Coordinate Equation of Parabola is given as the following

    \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F=0\)
  2. The General Parametric Equations of Parabola are given as

    \(x=A_1t^2 + B_1t +C_1,\hspace{.5cm}y=A_2t^2 + B_2t +C_2\)

    Given the General Parametric Equations of Parabola, the Co-effcients of Implicit Coordinate Equation of Parabola can be found out using the following
    1. Co-efficient of \(x^2\hspace{.3cm}A={A_2}^2\)   ...(1)
    2. Co-efficient of \(xy\hspace{.3cm}B=-2 A_1A_2\)   ...(2)
    3. Co-efficient of \(y^2\hspace{.3cm}C={A_1}^2\)   ...(3)
    4. Co-efficient of \(x\hspace{.3cm}D=B_2\begin{vmatrix}A_2 & B_2 \\ A_1 & B_1\end{vmatrix} - 2 A_2\begin{vmatrix}A_2 & C_2 \\ A_1 & C_1\end{vmatrix}\)   ...(4)
    5. Co-efficient of \(y\hspace{.3cm}E=B_1\begin{vmatrix}A_1 & B_1\\ A_2 & B_2\end{vmatrix} - 2 A_1\begin{vmatrix}A_1 & C_1 \\ A_2 & C_2\end{vmatrix}\)   ...(5)
    6. Constant of Equation \(F =\begin{vmatrix}A_1 & B_1\\ A_2 & B_2\end{vmatrix}\begin{vmatrix}C_1 & B_1\\ C_2 & B_2\end{vmatrix} + {\begin{vmatrix}A_1 & C_1 \\ A_2 & C_2\end{vmatrix}}^2\)   ...(6)
    The above formulae can also be used to find the Co-efficients of the General Parametric Equations of Parabola, when the Implicit Coordinate Equation of Parabola is given. The following gives the algorithm for the same
    1. From equation (1) given above we see that the Co-efficient of \(x^2\hspace{.3cm}A={A_2}^2\). Therefore the value of \(A_2\) can be one of \(\pm \sqrt{A}\).
    2. From equation (3) given above we see that the Co-efficient of \(y^2\hspace{.3cm}C={A_1}^2\). Therefore the value of \(A_1\) can be one of \(\pm \sqrt{C}\).
    3. From equation (2) given above we see that the Co-efficient of \(xy\hspace{.3cm}B=-2A_1A_2\). Therefore if the value of \(B < 0 \) then value of both \(A_1\) and \(A_2\) are either Positive or Negative. However, if the value of \(B > 0 \) then value of one of \(A_1\) and \(A_2\) is Negative and the other is Positive.

      Since we do not know what are the actual signs for \(A_1\) and \(A_2\), to do further calculations for the time being we assign arbitrary signage to these Co-efficients based on value of \(B\).
    4. Now, to find the value of Co-efficients Variables \(B_1\), \(B_2\), \(C_1\) and \(C_2\) we need 4 equations. However we only have 3 equations (equation (4), (5) and (6)), which provide the value of these Variables. Therefore we have to fix value of one of Co-efficients \(B_1\) or \(B_2\) as 0. The values of the remaining Co-efficient Variables are calculated as follows

      When value of \(B_2\) is fixed as 0, the value of \(B_1\), \(C_1\) and \(C_2\) are calculated as follows

      From equation (4) we have

      \(D=B_2\begin{vmatrix}A_2 & B_2 \\ A_1 & B_1\end{vmatrix} - 2 A_2\begin{vmatrix}A_2 & C_2 \\ A_1 & C_1\end{vmatrix}\)

      \(\Rightarrow D=B_2(A_2B_1-A_1B_2)-2A_2(A_2C_1-A_1C_2)\)

      Since \(B_2=0\) therefore

      \(D=-2A_2(A_2C_1-A_1C_2)\)

      \(\Rightarrow A_1C_2-A_2C_1=\frac{D}{2A_2}\)   ...(7)

      Now, From equation (5) we have

      \(E=B_1\begin{vmatrix}A_1 & B_1\\ A_2 & B_2\end{vmatrix} - 2 A_1\begin{vmatrix}A_1 & C_1 \\ A_2 & C_2\end{vmatrix}\)

      \(\Rightarrow E=B_1(A_1B_2-A_2B_1)-2A_1(A_1C_2-A_2C_1)\)

      Since \(B_2=0\) therefore

      \(E=-A_2{B_1}^2-2A_1(A_1C_2-A_2C_1)\)

      \(\Rightarrow A_1C_2-A_2C_1=-\frac{E+A_2{B_1}^2}{2A_1}\)   ...(8)

      Now, From equation (7) and (8) we have

      \(\frac{D}{2A_2}=-\frac{E+A_2{B_1}^2}{2A_1}\)

      \(\Rightarrow D\frac{A_1}{A_2}=-E-A_2{B_1}^2\)

      \(\Rightarrow {B_1}^2=-D\frac{A_1}{{A_2}^2}-\frac{E}{A_2}=-{\Large \frac{DA_1+EA_2}{{A_2}^2}}\)   ...(9)

      If the value of \({B_1}^2 > 0\) in equation (9) given above, then our assignment of sign for \(A_1\) and \(A_2\) in the previous was correct. However, if the value of \({B_1}^2 < 0\) in equation (9) given above, then we must multiply value of both \(A_1\) and \(A_2\) with -1 (i.e invert the sign of \(A_1\) and \(A_2\)) to provide them the correct sign. Also, the value of \(B_1\) can be found out as \(\sqrt{|{B_1}^2|}\).

      Now, From equation (6) we have

      \(F =\begin{vmatrix}A_1 & B_1\\ A_2 & B_2\end{vmatrix}\begin{vmatrix}C_1 & B_1\\ C_2 & B_2\end{vmatrix} + {\begin{vmatrix}A_1 & C_1 \\ A_2 & C_2\end{vmatrix}}^2\)

      \(\Rightarrow F =(A_1B_2-A_2B_1)(C_1B_2-C_2B_1) + {(A_1C_2-A_2C_1)}^2\)

      Since \(B_2=0\) therefore

      \(F =(-A_2B_1)(-C_2B_1) + {(A_1C_2-A_2C_1)}^2\)

      Now, Using equation (7) we have

      \(F =A_2C_2{B_1}^2 + {(\frac{D}{2A_2})}^2\)

      \(\Rightarrow C_2 = \frac{F - \frac{D^2}{4{A_2}^2}}{A_2{B_1}^2}={\Large \frac{4{A_2}^2F-D^2}{4{A_2}^3{B_1}^2}}\)   ...(10)

      Now, value of \(C_1\) can be found out by putting the value of \(C_2\) in equation (7) as

      \(A_1C_2-A_2C_1=\frac{D}{2A_2}\)

      \(\Rightarrow A_2C_1=A_1C_2-\frac{D}{2A_2}\)

      \(\Rightarrow C_1=C_2\frac{A_1}{A_2}-\frac{D}{2{A_2}^2}={\Large \frac{2A_1A_2C_2-D}{2{A_2}^2}}\)   ...(11)

      Similarly, when value of \(B_1\) is fixed as 0, the value of \(B_2\), \(C_1\) and \(C_2\) are calculated as follows

      From equation (5) we have

      \(E=B_1\begin{vmatrix}A_1 & B_1 \\ A_2 & B_2\end{vmatrix} - 2 A_1\begin{vmatrix}A_1 & C_1 \\ A_2 & C_2\end{vmatrix}\)

      \(\Rightarrow E=B_1(A_1B_2-A_2B_1)-2A_1(A_1C_2-A_2C_1)\)

      Since \(B_1=0\) therefore

      \(E=-2A_1(A_1C_2-A_2C_1)\)

      \(\Rightarrow A_1C_2-A_2C_1=-\frac{E}{2A_1}\)   ...(12)

      Now, From equation (4) we have

      \(D=B_2\begin{vmatrix}A_2 & B_2 \\ A_1 & B_1\end{vmatrix} - 2 A_2\begin{vmatrix}A_2 & C_2 \\ A_1 & C_1\end{vmatrix}\)

      \(\Rightarrow D=B_2(A_2B_1-A_1B_2)-2A_2(A_2C_1-A_1C_2)\)

      Since \(B_1=0\) therefore

      \(D=-A_1{B_2}^2-2A_2(A_2C_1-A_1C_2)\)

      \(\Rightarrow A_1C_2-A_2C_1=\frac{D+A_1{B_2}^2}{2A_2}\)   ...(13)

      Now, From equation (12) and (13) we have

      \(-\frac{E}{2A_1}=\frac{D+A_1{B_2}^2}{2A_2}\)

      \(\Rightarrow -E\frac{A_2}{A_1}=D+A_1{B_2}^2\)

      \(\Rightarrow {B_2}^2=-E\frac{A_2}{{A_1}^2}-\frac{D}{A_1}=-{\Large \frac{EA_2+DA_1}{{A_1}^2}}\)   ...(14)

      If the value of \({B_2}^2 > 0\) in equation (14) given above, then our assignment of sign for \(A_1\) and \(A_2\) in the previous was correct. However, if the value of \({B_2}^2 < 0\) in equation (14) given above, then we must multiply value of both \(A_1\) and \(A_2\) with -1 (i.e invert the sign of \(A_1\) and \(A_2\)) to provide them the correct sign. Also, the value of \(B_2\) can be found out as \(\sqrt{|{B_2}^2|}\).

      Now, From equation (6) we have

      \(F =\begin{vmatrix}A_1 & B_1\\ A_2 & B_2\end{vmatrix}\begin{vmatrix}C_1 & B_1\\ C_2 & B_2\end{vmatrix} + {\begin{vmatrix}A_1 & C_1 \\ A_2 & C_2\end{vmatrix}}^2\)

      \(\Rightarrow F =(A_1B_2-A_2B_1)(C_1B_2-C_2B_1) + {(A_1C_2-A_2C_1)}^2\)

      Since \(B_1=0\) therefore

      \(F =(A_1B_2)(C_1B_2) + {(A_1C_2-A_2C_1)}^2\)

      Now, Using equation (12) we have

      \(F =A_1C_1{B_2}^2 + {(-\frac{E}{2A_1})}^2\)

      \(\Rightarrow C_1 = \frac{F - \frac{E^2}{4{A_1}^2}}{A_1{B_2}^2}={\Large \frac{4{A_1}^2F-E^2}{4{A_1}^3{B_2}^2}}\)   ...(15)

      Now, value of \(C_2\) can be found out by putting the value of \(C_1\) in equation (12) as

      \(A_1C_2-A_2C_1=-\frac{E}{2A_1}\)

      \(\Rightarrow A_1C_2=A_2C_1-\frac{E}{2A_1}\)

      \(\Rightarrow C_2=C_1\frac{A_2}{A_1}-\frac{E}{2{A_1}^2}={\Large \frac{2A_1A_2C_1-E}{2{A_1}^2}}\)   ...(16)

  3. If both the Co-efficients of \(x^2\) (i.e. value of \(A\)) and \(y^2\) (i.e. value of \(C\)) in the Implicit Coordinate Equation of Parabola are Non-Zero, then 2 Sets of Parametric Equations can be found out for such Parabola (one by setting the value of \(B_1\) to 0 and other one by setting the value of \(B_2\) to 0).
    If the Co-efficient of \(x^2\) (i.e. value of \(A\)) in the Implicit Coordinate Equation of Parabola is Zero, then only one Set of Parametric Equations can be found out for such Parabola which will have \(B_1\) set to 0 (since the Co-efficient \(A_2\) will be equal to 0 and hence \(B_2\) cannot be zero) .
    Similarly, if the Co-efficient of \(y^2\) (i.e. value of \(C\)) in the Implicit Coordinate Equation of Parabola is Zero, then only one Set of Parametric Equations can be found out for such Parabola which will have \(B_2\) set to 0 (since the Co-efficient \(A_1\) will be equal to 0 and hence \(B_1\) cannot be zero).
Related Calculators
Parabola from Parametric Equation Calculator
Related Topics
Converting Parabola Equation from General Parametric to Implicit Coordinate,    Converting Parabola Equation from Standard Coordinate to Standard Parametric,    Converting Parabola Equation from Standard Parametric to Standard Coordinate,    Converting Parabola Equation from Explicit Coordinate to Parametric,    Converting Parabola Equation from Axis Aligned Parametric to Explicit/Implicit Coordinate,    Parametric Equations of Parabola,    Introduction to Parabola,    General Quadratic Equations in 2 Variables and Conic Sections
© Invincible IDeAS. All Rights Reserved