Lecture 2 [10 Feb, 2020] MATH 127 (Prof Farid)

MATH -127 (Coordinate Geometry)
Chapter-1 (Change of Axes)

Change of Axes (Transformation of Coordinates)

Shift the origin through parallel line:

1. Translation: 
    • General second-degree equation:  $$\begin{align*} & ax^2 + 2hxy + by^2 + 2hx + 2fy + c = 0 \\ & \frac{x^2 }{a^2} + \frac{y^2 }{b^2 } = 1 ~~ \textrm{or} ~~\frac{x^2 }{a^2} - \frac{y^2 }{b^2 } = 1 \\ & ax^2 + 2hxy + by^2 + 2gx + 2fy + c = ( a_1 x + b_1 y + c_1)~(a_2 x + b_2 y + c_2)\\ & y = ax^2 + bx + c \end{align*}$$  

If P is any point an ellipse, F and F' are two foci, then the relation between them is,

    •Shift the origin through parallel axes:
Algebra: $$\begin{align*} & x = x' + h\\ & y = y' + k \end{align*}$$  
there are transformation equation (remove the product term $2hxy$)

Fig: shifting the origin through parallel axes.

2. Rotation:  $\theta = \frac{1}{2} \tan^{-1} \left(\frac{2h}{a - b}\right)$  
    keeping the origin fixed, rotate the axis at an angle of $\theta$

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Solve Problems

    Question 1: Determine the equation of the curve  $2x^2 + 3 y^2 - 8x + 6y - 7 = 0$ , where the origin is shifted to the point (2,-1)

Solution: Given equation is,  $2x^2 + 3 y^2 - 8x + 6y - 7 = 0 \tag{1.1}\label{eq:1.1}$

replace $x$ by $x' + 2$ and $y$ by $y' + 1$ from equation (1.1) to get,

  $$\begin{align*} & 2(x' + 2)^2 + 3 (y' - 1 )^2 - 8(x' + 2 ) + 6(y' - 1 ) - 7 = 0 \\ \implies & 2 ( x'^2 + 4 x' + 4) + 3 (y'^2 - 2 y + 1) - 8x' - 16 + 6y' - 6 - 7 = 0 \\ \implies & 2 x'^2 + 8 x' + 8 + 3 y^2 - 6y + 3 - 8 x' - 16 + 6 y' - 13 = 0\\ \implies & 2x'^2 + 3 y'^2 - 18 = 0\\ \textrm{droping dashes we get,}\\ \implies & 2 x^2 + 3 y^2 = 18 \\ \implies &\frac{x^2 }{9 }+ \frac{y^2 }{6} = 1 \\ \implies &\frac{x^2}{3^2 } + \frac{y^2 }{(\sqrt{6})^2} = 1 \end{align*}$$  

Now, find the major axis, minor axis, foci.

    Question 2: Transform the equation  $17x^2 + 18xy - 7y^2 - 16x - 3y - 18 = 0$ to one in which there is no term containing x, y, xy

Solution: Given equation is,  $17x^2 + 18xy - 7y^2 - 16x - 3y - 18 = 0 \tag{2.1}\label{eq:2.1}$  

replace $x$ by $x' + 2$ and $y$ by $y' + 1$ from equaiton (2.1) to get,

  $$\begin{align*} & 17(x' + h)^2 + 18 (x' + h)(y' + k)- 7(y' + k)^2 - 16 (x' + h) - 32 (y' + k) - 18 = 0 \\ \implies & 17( x'^2 + 2 x'h + h^2 ) + 18(x'y' + x'k + y'h + hk) \\ &~~~~~~~~~~- 7 (y'^2 + 2y'k + k^2) - 16x' - 16h - 32y' - 32h - 18 = 0 \\ \implies & 17x^2 + 34 x' h + 17 h^2 + 18 s'y' + 18 x'k + 18y'h + 18hk \\ &~~~~~~~~~~- 7y^2 - 14y'k - 7k^2 - 16 x' - 16h - 32y' - 32k - 18 = 0 \\ \implies & 17x^2 + 18x'y' - 7y'^2 + 2(17h + 9k - 8)x' + (9h - 7k - 16)y \\ &~~~~~~~~~~+ 17h^2 + 18 hk - 7 k^2 - 16h - 32k - 18 = 0 \tag{2.3}\label{eq:2.3} \end{align*}$$  

to remove $x',y'$ from equation (2.3) we must consider, 

  $$\begin{align*} &17h + 9k - 8 = 0 ~~~\bigg| ~h = 1 \\ &9h - 7k - 16 = 0 ~~~\bigg| ~k = 1 \\ \end{align*}$$  

$\therefore$ The transformed equation is,  $17x^2 + 18xy - 7y^2 = 10 \tag{2.4}\label{eq:2.4}$  

Transformation equation for rotation:  $$\begin{align*} & \theta = \frac{1}{2} \tan^{-1} \left(\frac{2h}{a - b}\right)\\ &x = x' \cos \theta - y' \sin \theta \\ &y = x'\sin \theta + y'\cos \theta \end{align*}$$  

To remove $xy$ from equation (2.4), substitute $x$ by $x'\cos \theta - y'\sin \theta$ and $y$ by $x'\sin \theta + y' \cos \theta$ to get,

  $$\begin{align*} & 17(x'\cos\theta - y' \sin\theta)^2 + 8 ( x'\cos\theta - y'\sin\theta)(x'\sin\theta + y'\cos\theta)\\ & ~~~~~~~~~~ 7(x'\sin \theta + y'\cos \theta)^2 = 10 \\ \implies & 17(x'^2 \cos^2\theta - 2 x' y^2 \sin\theta\cos\theta + y'^2 \sin^2 \theta) + 18(x'^2 \sin \theta \cos \theta + x'y' \sin \theta \cos \theta \\ &~~~~~~~~~~ x'y'\sin^2 \theta - y'^2 \sin \theta \cos \theta) - 7 ( x'^2 \sin^2 \theta + 2 x'y' \sin \theta \cos \theta + y'^2 \cos \theta) = 10\\ \implies &17 x'^2 \cos^2 \theta - 34 x'y' \sin \theta \cos \theta + 17 y^2 \sin^2 \theta + 18 x'^2 \sin \theta \cos \theta + 18x'y' \sin \theta \cos \theta \\ &~~~~~~~~~~ - 18x'y' \sin^2 \theta - 18 y^2 \sin \theta \cos \theta - 7 x'^2 \sin^2 \theta - 14 y'y\sin \theta\cos \theta - 7 y^2 \cos^2 \theta = 0 \\ \implies & (17 \cos^2 \theta + 18 \sin \theta \cos \theta - 7 \sin^2 \theta )x'^2 - ( 34 \sin \theta \cos\theta + 14 \sin\theta \cos\theta- 18 \sin^2 \theta \\ &~~~~~~~~~~ - 14\sin\theta \cos\theta)y'^2 + (17\sin^2 \theta - 18 \sin\theta\cos\theta - 7 \cos^2\theta)y^2 = 0 \end{align*}$$  

To remove the product term, the factor must be zero. Therefore to remove $x'y'$ from(2.5) we get,

  $$\begin{align*} &18 (\cos^2 \theta - \sin^2 \theta) = 48\sin\theta\cos\theta \\ \implies & 18\cos2\theta = 25 \sin2\theta\\ \implies &\tan 2 \theta = \frac{3}{4}\\ \implies &\theta = \frac{1}{2}\tan^{-1} \left(\frac{3}{4}\right) ~~~~~ \textrm{Here,}~~ h = \frac{3}{2} ~\textrm{and}~ a - b = 4 \end{align*}$$