The two equations are:
The points where the first meets the second are the values of x and y that simultaneously satisfy both equations:
Substitute using (1) for x in (2) and solve the formed quadratic:
x² + 4y² = 4
→ (2 - 2y)² + 4y² = 4
→ 4 - 8y + 4y² + 4y² -4 = 0
→ 8y² - 8y = 0
→ 8y(y - 1) = 0
→ y = 0 or y = 1
Substituting these values back into (1) gives:
y = 0 → x = 2 - 2× 0 = 2 - 0 = 2
y = 1 → x = 2 - 2× 1 = 2 - 2 = 0
→ The line meets the curve at the points (2, 0) and (0, 1)
The points (0, 1) and (2, 0).
The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))
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)
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)
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)
If: x -y = 2 then x = 2 +y If: x^2 -4y^2 = 5 then (2 +y)^2 -4y^2 = 5 So: 4+4y +y^2 -4y^2 = 5 Collecting like terms: 4y -3y^2 -1 = 0 Solving the above quadratic equation: y = 1 or y = 1/3 Therefore by substitution the points of contact are at: (3, 1) and (7/3, 1/3)
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)
0
If: 2y-x = 0 Then: 4y^2 = x^2 If: x^2 +y^2 = 20 Then: 4y^2 +y^2 = 20 So: 5y^2 = 20 Dividing both sides by 5: y^2 = 4 Square root of both sides: y = - 2 or + 2 By substituting points of contact are at: (4, 2) and (-4, -2)