when the origin is shifted to the point (2,3)the transformed equation of a curve is x²+3xy-2y²+17x-7y-11=0.find the original equation of the curve
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Given when the origin is shifted to the point (2,3)the transformed equation of a curve is x²+3xy-2y²+17x-7y-11 = 0.find the original equation of the curve
- Given transformed equation will be
- X^2 + 3XY – 2Y^2 + 17 X – 7Y – 11 = 0
- X = x – h and Y = y – k
- Given new origin = (2,3)
- So we get
- X = x – 2 and Y = y – 3
- Substituting X and Y we get
- (x – 2)^2 + 3 (x – 2) (y – 3) – 2 (y – 3)^2 + 17 (x – 2) – 7(y – 3) – 11 = 0 x^2 + 4 – 4x + 3(xy – 2y – 3x + 6) – 2(y^2 + 9 – 6y) + 17x – 34 – 7y + 21 – 11 = 0
- x^2 + 4 – 4x + 3xy – 6y – 9x + 18 – 2y^2 – 18 + 12y + 17x – 34 – 7y + 21 – 11 = 0
- x^2 + 3xy – 2y^2 + 4x – y – 20 = 0 will be the original equation. of curve.
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https://brainly.in/question/25977903
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sinstitute XAnd Y hope it helps you
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