# The points of intersection of the line y=x-1 and the curve `2x^2 -2xy+y^2=13` are (2,-1) or (3.6,3.8).If these points of intersection also lie on the curve y=`(x-a)^2` +b,where a and b are...

The points of intersection of the line y=x-1 and the curve `2x^2 -2xy+y^2=13` are (2,-1) or (3.6,3.8).If these points of intersection also lie on the curve y=`(x-a)^2` +b,where a and b are constants,find the value of a and b.Thanks

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### 1 Answer

The points of intersection of the line y = x - 1 and the curve `2x^2 - 2xy + y^2 = 13` are given as (2, -1) and (3.6, 3.8).

It can be seen that for the point (2, -1), 2 - 1 = 1 `!=` -1. Also, for the point (3.6, 3.8), 3.6 - 1 = 2.6 `!=` 3.8. The given points do not lie on the line y = x - 1. As a result they cannot be the point of intersection of the line y = x - 1 and the curve 2x^2 - 2xy + y^2 = 13.

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The points of intersection of the line y=3x-7 and the curve are (2,-1) or (3.6,3.8).If these points of intersection also lie on the curve y=+b,where a and b are constants,find the value of a and b.Thanks