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What are the points of contact when the line x -y equals 2 meets the curve x squared -4y squared equals 5?
Wiki User
∙ 7y ago
If: x-y = 2 then x^2 = (2+y)^2
If: x^2 -4y^2 = 5 then x^2 = 4y^2 +5
So: 4y^2 +5 = (2+y)^2
Expanding brackets and transposing terms: 3y^2 +1 -4y = 0
Factorizing: (3y-1)(y-1) = 0 => y = 1/3 or y = 1
Therefore by substitution points of contact are at: (7/3, 1/3) and (3, 1)
Wiki User
∙ 7y ago
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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 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)
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))
What are the points of intersection of the line y equals -8-3x with the curve y equals -2-4x-x squared?
They work out as: (-3, 1) and (2, -14)
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)
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 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 is the value of k when y equals 3x plus 1 is a line tangent to the curve of x squared plus y squared equals k hence finding the point of contact of the line to the curve?
It is (-0.3, 0.1)
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)
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))
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 points of intersection of the line y equals -8-3x with the curve y equals -2-4x-x squared?
They work out as: (-3, 1) and (2, -14)
What is the value of c when y equals x plus c is a tangent to the curve y equals 3 -x -5x squared hence finding the point of contact of the line with the curve showing work?
-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)
What are points of intersection of the line 3x -y equals 5 with the curve of 2x squared plus y squared equals 129?
Straight line: 3x-y = 5 Curved parabola: 2x^2 +y^2 = 129 Points of intersection works out as: (52/11, 101/11) and (-2, -11)
At what point is the line of y equals x -4 tangent to the curve of x squared plus y squared equals 8?
(2, -2)
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)
