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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 else can I help you with?
What does the function y equals x squared look like?
y = x2 describes a parabolic curve with a focal point at the location 0, 0 and an infinite range greater than or equal to zero.
Which term best describes the point line or curve defined by the intersection of a cone and a plane?
It is the base of the cone
What is a curve?
A geometrical curve is defined as any set of points. Therefore, counter-intuitively, a straight line is also a geometrical curve.
What is the curve between 2 points in a circle called?
If the curve is part of the circumference of the circle, it is called an arc.
What is the equation to the tangent line to the curve xy plus y2 equals 0 at the point x equals 3 and y equals 0?
y=0. note. this is a very strange "curve". If y=0 then any value of x satisfies the equation, leading to a curve straight along the y axis. For any non-zero value of y the curve simplifies to y = -x. The curve is not differentiable at the origin.
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 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)
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)
