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What are the points of contact that the line of 2x plus y equals 5 makes with the curve of x squared -y squared equals 3 showing work?
Wiki User
∙ 7y ago
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)
Wiki User
∙ 7y ago
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What are the full range of values for k when the line y equals x -3 meets the curve of x squared -3y squared equals k showing work?
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What are the points of intersection of the line x -y equals 2 with x squared -4y squared equals 5?
The points of intersection are: (7/3, 1/3) and (3, 1)
What are the points of contact that the curve of y equals x squared -x -12 makes with the x and y axes on the Cartesian plane?
If: y = x2-x-12 Then points of contact are at: (0, -12), (4, 0) and (-3, 0)
What are the points of intersection of 3x -2y -1 equals 0 with 3x squared -2y squared equals -5 on the Cartesian plane?
Points of intersection work out as: (3, 4) and (-1, -2)
Where are the points of contact when the line 3x -y equals 5 passes through the curve 2x squared plus y squared equals 129?
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)
What are the points of contact between the line 3x -2y equals 1 and the curve 3x squared -2y squared plus 5 equals 0?
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)
What are the full range of values for k when the line y equals x -3 meets the curve of x squared -3y squared equals k showing work?
0
What are the points of intersection of the line x -y equals 2 with x squared -4y squared equals 5?
The points of intersection are: (7/3, 1/3) and (3, 1)
What are the points of contact that the curve of y equals x squared -x -12 makes with the x and y axes on the Cartesian plane?
If: y = x2-x-12 Then points of contact are at: (0, -12), (4, 0) and (-3, 0)
What are the points of intersection of 3x -2y -1 equals 0 with 3x squared -2y squared equals -5 on the Cartesian plane?
Points of intersection work out as: (3, 4) and (-1, -2)
Where are the points of contact when the line 3x -y equals 5 passes through the curve 2x squared plus y squared equals 129?
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)
What are the points of contact when the line 2x plus 5y equals 4 transverses the curve y squared equals x plus 4?
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)
What are the possible points of contact when the line 2y -x equals 0 meets the curve x squared plus y squared equals 20?
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)
Where are the points of contact when the line 2x plus 5 equals 4 crosses the curve y squared equals x plus 4 on the Cartesian plane?
The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))
Where are the points of intersection of the equations 4y squared -3x squared equals 1 and x -2y equals 1?
The points of intersection of the equations 4y^2 -3x^2 = 1 and x -2 = 1 are at (0, -1/2) and (-1, -1)
What are the points of intersection of x plus y equals 7 with x squared plus y squared equals 25 showing key stages of work?
If: x+y = 7 and x2+y2 = 25 Then: x = 7-y and so (7-y)2+y2 = 25 => 2y2-14y+24 = 0 Solving the quadratic equation: y = 4 and y = 3 By substitution points of intersection: (3, 4) and (4, 3)
What are the points of contact between the line 2y -x equals 0 and the curve x squared plus y squared equals 20?
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)
