0
What are points of intersection of the line 3x -y equals 5 with the curve of 2x squared plus y squared equals 129?
Wiki User
∙ 8y ago
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)
Wiki User
∙ 8y ago
Add your answer:
Earn +20 pts
What is mean by first order continuity curve?
The first order continuity curve is a term used in geometry to describe parametric first derivatives that are in proportion at the intersection on at least two successive sections of the curve.
Difference between integral and differential calculus?
Differential calculus is concerned with finding the slope of a curve at different points. Integral calculus is concerned with finding the area under a curve.
Right circular cone is intersected by a plane that is parallel to one edge of the cone?
I'm assuming you are looking for the name of the conic section produced by this type of intersection? If a right circular cone is intersected by a plane parallel to one edge of the cone, the resulting curve of intersection would be a parabola. If the intersecting plane was parallel to the base, it would be a circle. If the intersecting plane was at any angle between being parallel to the base and being parallel to an edge, it would produce an ellipse or part of an ellipse (depending on whether the intersection was completely within the cone).
Is curve an action verb?
curve is an action verb
Is the word curve a verb?
The word curve can be used as either a verb or a noun. As a verb: when you throw a ball, its path will curve downward, because of gravity. As a noun: the equation can be drawn on the graph as a smooth curve.
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)
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)
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 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 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)
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)
