If: 2x+y = 5 then y = 5-x
If: x^2 - y^2 = 3 then x^2 -(5-x)^2 = 3
So: x^2 -(25 -20x +4x^2) = 3
Removing the brackets and subtracting 3 from both sides: -3x^2-28+20x = 0
Using the quadratic equation formula: x = 14/3 or x = 2
Therefore by substitution points of contact are at: (14/3, -13/3) and (2, 1)
Chat with our AI personalities
The points of intersection are: (7/3, 1/3) and (3, 1)
0
If: y = x2-x-12 Then points of contact are at: (0, -12), (4, 0) and (-3, 0)
Points of intersection work out as: (3, 4) and (-1, -2)
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)