Let x2−2xy−3y2=4 Then find the range of 2x2−2xy+y2

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Jazlyn Mitchell · Accepted Answer

Others have pointed out where you went wrong. I just want to provide an alternative proof.x2−2xy−3y2=4=(x+y)(x−3y)Denote z=x+y,w=x−3y, thenzw=4,x=3z+w4,y=z−w4We want to find the range of2x2−2xy+y2=116(5w2+14wz+13z2)=72+116(5w2+13z2)tag1 (1)under the constraint zw=4.Clearly (1) can be as large as you want. And if you apply AM-GM you get the minimum72+116(5w2+13z2)≥72+1162513|wz|=7+652

slokningol09 · Accepted Answer

Consider the curves x2−2xy−3y2=4 (C) and 2x2−2xy+y2=k2+4 (E) At the points where they intersect, we can substitute the value of -2xy from the first into the second to getx2+4y2=k2 (E)This represents an ellipse centered at the origin. Now, either with a bit of graph sketching, or by considering (C) as a quadratic in x2, deduce that y can range anywhere in the real numbers, and that (C) is a hyperbola. Increasing k in (E) just enlarges the ellipse, and it’s not hard to see that there is no upper bound on k for (E) and (C) to intersect. For obtaining a lower bound, we need to find k such that the two curves touch. Note that if (x,y) lies on the two curves, then so does (-x,-y). So we’re looking for exactly two intersections. Then substituting x=±k2−4y2 in (C) gives a quadratic in y2:65y4−y2(18k2−56)+(k2−4)2=0Setting the discriminant equal to zero will give k2=65−12 and hence we have the desired range:65+72≤2x2−2xy+y2=k2+4&amp;lt;∞

Let \(\displaystyle{x}^{{{2}}}-{2}{x}{y}-{3}{y}^{{{2}}}={4}\) Then find the range of

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