Loading [MathJax]/jax/output/CommonHTML/jax.js

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:  ax2+2hxy+by2+2hx+2fy+c=0x2a2+y2b2=1  or  x2a2y2b2=1ax2+2hxy+by2+2gx+2fy+c=(a1x+b1y+c1) (a2x+b2y+c2)y=ax2+bx+c  

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:x=x+hy=y+k  
there are transformation equation (remove the product term 2hxy)






Fig: shifting the origin through parallel axes.



2. Rotation:  θ=12tan1(2hab)  
    keeping the origin fixed, rotate the axis at an angle of θ







 


Solve Problems

    Question 1: Determine the equation of the curve  2x2+3y28x+6y7=0 , where the origin is shifted to the point (2,-1)

Solution: Given equation is,  2x2+3y28x+6y7=0

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

  2(x+2)2+3(y1)28(x+2)+6(y1)7=02(x2+4x+4)+3(y22y+1)8x16+6y67=02x2+8x+8+3y26y+38x16+6y13=02x2+3y218=0droping dashes we get,2x2+3y2=18x29+y26=1x232+y2(6)2=1  

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


    Question 2: Transform the equation  17x2+18xy7y216x3y18=0 to one in which there is no term containing x, y, xy

Solution: Given equation is,  17x2+18xy7y216x3y18=0  

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

  17(x+h)2+18(x+h)(y+k)7(y+k)216(x+h)32(y+k)18=017(x2+2xh+h2)+18(xy+xk+yh+hk)          7(y2+2yk+k2)16x16h32y32h18=017x2+34xh+17h2+18sy+18xk+18yh+18hk          7y214yk7k216x16h32y32k18=017x2+18xy7y2+2(17h+9k8)x+(9h7k16)y          +17h2+18hk7k216h32k18=0  

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

  17h+9k8=0   | h=19h7k16=0   | k=1  

The transformed equation is,  17x2+18xy7y2=10  

Transformation equation for rotation:  θ=12tan1(2hab)x=xcosθysinθy=xsinθ+ycosθ  

To remove xy from equation (2.4), substitute x by xcosθysinθ and y by xsinθ+ycosθ to get,

  17(xcosθysinθ)2+8(xcosθysinθ)(xsinθ+ycosθ)          7(xsinθ+ycosθ)2=1017(x2cos2θ2xy2sinθcosθ+y2sin2θ)+18(x2sinθcosθ+xysinθcosθ          xysin2θy2sinθcosθ)7(x2sin2θ+2xysinθcosθ+y2cosθ)=1017x2cos2θ34xysinθcosθ+17y2sin2θ+18x2sinθcosθ+18xysinθcosθ          18xysin2θ18y2sinθcosθ7x2sin2θ14yysinθcosθ7y2cos2θ=0(17cos2θ+18sinθcosθ7sin2θ)x2(34sinθcosθ+14sinθcosθ18sin2θ          14sinθcosθ)y2+(17sin2θ18sinθcosθ7cos2θ)y2=0  

To remove the product term, the factor must be zero. Therefore to remove xy from(2.5) we get,

  18(cos2θsin2θ)=48sinθcosθ18cos2θ=25sin2θtan2θ=34θ=12tan1(34)     Here,  h=32 and ab=4  


     

Post a Comment

Post a Comment (0)

Previous Post Next Post