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What are the points of intersection of the line of x -2y equals 1 with the curve of 3xy -y squared equals 8 showing work?
If: x-2y = 1 and 3xy-y2 = 8 Then: x =1+2y and so 3(1+2y)y-y2 = 8 => 3y+5y2-8 = 0 Solving the quadratic equation: y = 1 or y = -8/5 Points of intersection by substitution: (3, 1) and (-11/5, -8/5)
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 intersection of the line 3x -y equals 5 with the curve of 2x squared plus y squared equals 129 with key stages of work shown?
If: 3x-y = 5 and 2x2+y2 = 129 Then: x = 5/3+y/3 and so 2(5/3+y/3)2+y2 = 129 => 50+20y+2y2+9y2 = 1161 Therefore: 11y2+20y-1111 = 0 Solving the quadratic equation: y = -11 or y = 101/11 By substitution points of intersection are: (-2, -11) and (52/11, 101/11)
What are the points of intersection of the line y equals 2x plus 1 with the curve 5y squared plus 5x squared equals 1 showing work?
5y2 + 5x2 = 1 Substitute y = 2x + 1 into this equation: 5*(2x + 1)2 + 5x2 = 1 5*(4x2 + 4x + 1) + 5x2 - 1 = 0 25x2 + 20x + 4 = 0 which factorises to (5x + 2)2 = 0 Therefore x = -2/5 And then y = 2x + 1 gives y = 1/5 So the two points of intersection are coincident, at (-2/5, 1/5)
What is the length of the line y equals 8 -x that meets the curve of y equals x squared plus 4x plus 2 at two distinctive points?
7*sqrt(2) = 9.899 to 3 dp
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)
Where are the points of intersection of the straight line 3x -y equals 5 with the curve 2x squared plus y squared equals 129?
(52/11, 101/11) and (-2, -11) Rearrange 3x-y = 5 into y = 3x-5 and substitute this into the curve equation and then use the quadratic equation formula to find the values of x which leads to finding the values of y by substituting the values of x into y = 3x-5.
What are the points of intersection of the line of x -2y equals 1 with the curve of 3xy -y squared equals 8 showing work?
If: x-2y = 1 and 3xy-y2 = 8 Then: x =1+2y and so 3(1+2y)y-y2 = 8 => 3y+5y2-8 = 0 Solving the quadratic equation: y = 1 or y = -8/5 Points of intersection by substitution: (3, 1) and (-11/5, -8/5)
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 intersection when the line y equals 10x -12 crosses the curve y equals x squared plus 20x plus 12?
If: y = 10x -12 and y = x^2 +20x +12 Then: x^2 +20x +12 = 10x -12 Transposing terms: x^2 +10x +24 = 0 Factorizing: (x+6)(x+4) = 0 => x = -6 or x = -4 Points of intersection by substitution are at: (-6, -72) and (-4, -52)
What are the points of intersection when the line y equals -8 -3x passes through the curve y equals -2 -4x - x squared?
If: y = -8 -3x and y = -2 -4x -x^2 Then: -8 -3x = -2 -4x - x^2 Transposing terms: x^2 +x -6 = 0 Factorizing: (x-2)(x+3) = 0 => x = 2 or x = -3 Points of intersection by substitution are at: (2, -14) and (-3, 1)
What are the points of intersection of the line 3x -y equals 5 with the curve of 2x squared plus y squared equals 129 with key stages of work shown?
If: 3x-y = 5 and 2x2+y2 = 129 Then: x = 5/3+y/3 and so 2(5/3+y/3)2+y2 = 129 => 50+20y+2y2+9y2 = 1161 Therefore: 11y2+20y-1111 = 0 Solving the quadratic equation: y = -11 or y = 101/11 By substitution points of intersection are: (-2, -11) and (52/11, 101/11)
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 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)
Where are the points of intersection on the curve and line of x squared plus y squared plus 4x plus 6y minus 40 equals 0 and x equals 10 plus y?
x2+y2+4x+6y-40 = 0 and x = 10+y Substitute the second equation into the first equation: (10+y)2+y2+4(10+y)+6y-40 = 0 2y2+30y+100 = 0 Divide all terms by 2: y2+15y+50 = 0 (y+10)(y+5) = 0 => y = -10 or y = -5 Substitute the above values into the second equation to find the points of intersection: Points of intersection are: (0, -10) and (5, -5)
What is the length of the line y equals 8 -x that meets the curve of y equals x squared plus 4x plus 2 at two distinctive points?
7*sqrt(2) = 9.899 to 3 dp
What are the points of intersection when the line 3x -y equals 5 crosses the curve 2x squared plus y squared equals 129?
If: 3x-y = 5 Then: y^2 = 9x^2 -30x +25 If: 2x^2 +y^2 = 129 Then: 2x^2 +9x^2 -30x +25 = 129 Transposing terms: 11x^2 -30x -104 = 0 Factorizing the above: (11x-52)(x+2) = 0 meaning x = 52/11 or x = -2 Therefore by substitution points of intersection are at: (52/11, 101/11) and (-2, -11)
