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 x2a2−y2b2=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)
2. Rotation:
θ=12tan−1(2ha−b)
keeping the origin fixed, rotate the axis at an angle of θ
Solve Problems
Question 1: Determine the equation of the curve 2x2+3y2−8x+6y−7=0 , where the origin is shifted to the point (2,-1)
Solution: Given equation is, 2x2+3y2−8x+6y−7=0
replace x by x′+2 and y by y′+1 from equation (1.1) to get,
2(x′+2)2+3(y′−1)2−8(x′+2)+6(y′−1)−7=0⟹2(x′2+4x′+4)+3(y′2−2y+1)−8x′−16+6y′−6−7=0⟹2x′2+8x′+8+3y2−6y+3−8x′−16+6y′−13=0⟹2x′2+3y′2−18=0droping dashes we get,⟹2x2+3y2=18⟹x29+y26=1⟹x232+y2(√6)2=1
Now, find the major axis, minor axis, foci.
Question 2: Transform the equation 17x2+18xy−7y2−16x−3y−18=0 to one in which there is no term containing x, y, xy
Solution: Given equation is, 17x2+18xy−7y2−16x−3y−18=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)2−16(x′+h)−32(y′+k)−18=0⟹17(x′2+2x′h+h2)+18(x′y′+x′k+y′h+hk) −7(y′2+2y′k+k2)−16x′−16h−32y′−32h−18=0⟹17x2+34x′h+17h2+18s′y′+18x′k+18y′h+18hk −7y2−14y′k−7k2−16x′−16h−32y′−32k−18=0⟹17x2+18x′y′−7y′2+2(17h+9k−8)x′+(9h−7k−16)y +17h2+18hk−7k2−16h−32k−18=0
to remove x′,y′ from equation (2.3) we must consider,
17h+9k−8=0 | h=19h−7k−16=0 | k=1
∴ The transformed equation is, 17x2+18xy−7y2=10
Transformation equation for rotation: θ=12tan−1(2ha−b)x=x′cosθ−y′sinθy=x′sinθ+y′cosθ
To remove xy from equation (2.4), substitute x by x′cosθ−y′sinθ and y by x′sinθ+y′cosθ to get,
17(x′cosθ−y′sinθ)2+8(x′cosθ−y′sinθ)(x′sinθ+y′cosθ) 7(x′sinθ+y′cosθ)2=10⟹17(x′2cos2θ−2x′y2sinθcosθ+y′2sin2θ)+18(x′2sinθcosθ+x′y′sinθcosθ x′y′sin2θ−y′2sinθcosθ)−7(x′2sin2θ+2x′y′sinθcosθ+y′2cosθ)=10⟹17x′2cos2θ−34x′y′sinθcosθ+17y2sin2θ+18x′2sinθcosθ+18x′y′sinθcosθ −18x′y′sin2θ−18y2sinθcosθ−7x′2sin2θ−14y′ysinθcosθ−7y2cos2θ=0⟹(17cos2θ+18sinθcosθ−7sin2θ)x′2−(34sinθcosθ+14sinθcosθ−18sin2θ −14sinθcosθ)y′2+(17sin2θ−18sinθcosθ−7cos2θ)y2=0
To remove the product term, the factor must be zero. Therefore to remove x′y′ from(2.5) we get,
18(cos2θ−sin2θ)=48sinθcosθ⟹18cos2θ=25sin2θ⟹tan2θ=34⟹θ=12tan−1(34) Here, h=32 and a−b=4
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