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Question 14 of 16

Q.When the origin is shifted to the point (2,3)(2, 3) the transformed equation of a curve is x2+3xy−2y2+17x−7y−11=0x^2 + 3xy - 2y^2 + 17x - 7y - 11 = 0. Find the original equation of the curve.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2025Subjective· 4mImportance★★★★★
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Since the origin was shifted to (2,3)(2,3), new and old coordinates relate by X=x−2, Y=y−3X=x-2,\ Y=y-3; substituting into the transformed equation and simplifying recovers the original equation.

When the origin is shifted to (h,k)=(2,3)(h,k)=(2,3), a point's old coordinates (x,y)(x,y) and new coordinates (X,Y)(X,Y) are related by x=X+h, y=Y+kx=X+h,\ y=Y+k, i.e. X=x−2, Y=y−3X=x-2,\ Y=y-3.

The transformed equation is X2+3XY−2Y2+17X−7Y−11=0X^2+3XY-2Y^2+17X-7Y-11=0. Substitute X=x−2, Y=y−3X=x-2,\ Y=y-3:

(x−2)2+3(x−2)(y−3)−2(y−3)2+17(x−2)−7(y−3)−11=0(x-2)^2+3(x-2)(y-3)-2(y-3)^2+17(x-2)-7(y-3)-11=0

Expand each term:

(x−2)2=x2−4x+4(x-2)^2 = x^2-4x+4

3(x−2)(y−3)=3(xy−3x−2y+6)=3xy−9x−6y+183(x-2)(y-3) = 3(xy-3x-2y+6) = 3xy-9x-6y+18

−2(y−3)2=−2(y2−6y+9)=−2y2+12y−18-2(y-3)^2 = -2(y^2-6y+9) = -2y^2+12y-18

17(x−2)=17x−3417(x-2) = 17x-34

−7(y−3)=−7y+21-7(y-3) = -7y+21

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