Points of intersection work out as: (3, 4) and (-1, -2)
The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))
The points of intersection are: (7/3, 1/3) and (3, 1)
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)
Improved Answer:-If: 2x+y = 5 and x^2 -y^2 = 3Then by rearranging: y = 5 -2x and -3x^2 -28+20x = 0Solving the above quadratic equation: x = 2 and x = 14/3By substitution points of intersection are: (2, 1) and (14/3, -13/3)
Points of intersection work out as: (3, 4) and (-1, -2)
The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))
The points are (-1/3, 5/3) and (8, 3).Another Answer:-The x coordinates work out as -1/3 and 8Substituting the x values into the equations the points are at (-1/3, 13/9) and (8, 157)
Equations: 3x-2y = 1 and 3x^2 -2y^2 +5 = 0 By combining the equations into one single quadratic equation and solving it the points of contact are made at (3, 4) and (-1, -2)
The points of intersection are: (7/3, 1/3) and (3, 1)
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)
14
If 2x+5y = 4 and y^2 = x+4 then by combining the equations into a single quadratic equation and solving it the points of contact are made at (12, -4) and (-7/2, 3/2)
Improved Answer:-If: 2x+y = 5 and x^2 -y^2 = 3Then by rearranging: y = 5 -2x and -3x^2 -28+20x = 0Solving the above quadratic equation: x = 2 and x = 14/3By substitution points of intersection are: (2, 1) and (14/3, -13/3)
If: 2y-x = 0 then 4y^2 = x^2 So : 4y^2 +y^2 = 20 or 5y^2 = 20 or y^2 = 4 Square rooting both sides: y = -2 or y = 2 Therefore possible points of contact are at: (4, 2) and (-4. -2)
The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
Parabola: y = 6x^2 -7x+2 It makes contact with the y axis at: (0, 2) It makes contact with the x axis at: (2/3, 0) and (1/2, 0)