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Find a point at which origin is shifted such that transformed equation of 2x^(2)+y^(2)-12xy+16=0 has no term containing x and constant term. |
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Answer» Solution :Let origin be shifted to the point `(h,k)`. `:. "Put" x=X+h` and `y=Y+km` `2(X+h)^(2)+(Y+k)^(2)-12(X+h)+(Y+k)+16=0` `impliesX^(2)+2h^(2)+4hX+Y^(2)+k^(2)+2kY-12X-12h+Y+k+16=0` `implies 2X^(2)+Y^(2)+X(4h-12)+Y(2k+1)+(2h^(2)+k^(2)-12h+k+16)-0`……`(1)` The EQUATION will be INDEPENDENT of `X` if `4h-12=0` `implies h=3` The equation will be independent of CONSTANT term if `2h^(2)+k^(2)-12h+k+16=0` `implies18+k^(2)-36+k+16=0` `k^(2)+k-2=0` `implies (k+2)(k-1)=0` `implies k=-2` or `k=1` `:.` Required point is `(3,-2)` or `(3,1)` |
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