If: x -y = 2 then x^2 = (2+y)^2 => 4+4y+y^2
If: x^2 -4y^2 = 5 then x^2 = 5+4y^2
So: 5+4y^2 = 4+4y+y^2
Transposing terms: 3y^2 -4y +1 = 0
Factorizing the above: (3y-1)(y-1) = 0 meaning y = 1/3 or y = 1
By substitution contacts are made at: (7/3, 1/3) and (3, 1)
The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))
5.477225575 squared equals 30.
The number that equals 121 when squared is 11.
0
b = sqrt32 or 4 root 2
The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))
-2
It is (-0.3, 0.1)
5.477225575 squared equals 30.
The number that equals 121 when squared is 11.
0
b = sqrt32 or 4 root 2
No, it equals -2xy. lrn2math
It works out that line 3x-y = 5 makes contact with the curve 2x^2 +y^2 = 129 at (52/11, 101/11) and (-2, -11)
If: y = x^2 -10x +13 and y = x^2 -4x +7 Then: x^2 -10x +13 = x^2 -4x +7 Transposing terms: -6x +6 = 0 => -6x = -6 => x = 1 Substituting the value of x into the original equations point of contact is at: (1, 4)
4
C equals the square root of 1000 or 31.622776601683793319988935444327...