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How do you prove in 3 different ways that the line of y equals x -4 is tangent to the curve of x squared plus y squared equals 8?
If: y = x-4 and y = x2+y2 = 8 then 2x2-8x+8 = 0 and the 3 ways of proof are:
1 Plot the given values on a graph and the line will touch the curve at one point
2 The discriminant of b2-4ac of 2x2-8x+8 must equal 0
3 Solving the equation gives x = 2 or x = 2 meaning the line is tangent to the curve
What else can I help you with?
What is the value of k when y equals 3x plus 1 and is a line tangent to the curve of x squared plus y squared equals k?
k = 0.1
What is the value of y when y equals 2x plus 1.25 is a tangent to the curve y squared equals 10x?
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
What is the definition of a tangent line?
A tangent is a line that just touches a curve at a single point and its gradient equals the rate of change of the curve at that point.
What is the value of k when the straight y equals 3x-1 is a tangent to the curve x squared plus y squared equals k?
2
How can you prove that the straight line of y equals x plus k is a tangent to the curve of x squared plus y squared equals 4 and only when k squared equals 8?
You don't.Furthermore, how would you prove your not a virgin? Either way your fu'd.Improved Answer:-y = x+kx2+y2 = 4Substitute y = x+k into the bottom equation:x2+(x+k)(x+k) = 4x2+x2+2kx+k2 = 4k2 = 8 so therefore:2x2+2kx+8-4 = 0 => 2x2+2kx+4 = 0For the straight line to be a tangent to the curve the discriminant b2-4ac of the quadratic equation must equal zero:Hence:-(2*k)2-4*2*4 = 0k2 = 8So: 4*8-4*2*4 = 0 => 32-32 = 0Therefore the straight line y = x+k is a tangent to the curve x2+y2 = 4
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 and is a line tangent to the curve of x squared plus y squared equals k?
k = 0.1
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 value of y when y equals 2x plus 1.25 is a tangent to the curve y squared equals 10x?
If the line y = 2x+1.25 is a tangent to the curve y^2 = 10x then it works out that when x = 5/8 then y = 5/2
What is the value of k in the line of y equals kx plus 1 and is tangent to the curve of y squared equals 8x?
If: y = kx+1 is a tangent to the curve y^2 = 8x Then k must equal 2 for the discriminant to equal zero when the given equations are merged together to equal zero.
What is the definition of a tangent line?
A tangent is a line that just touches a curve at a single point and its gradient equals the rate of change of the curve at that point.
What is the value of k when the straight y equals 3x-1 is a tangent to the curve x squared plus y squared equals k?
2
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 is the point of contact when the tangent line y equals x plus c touches the curve y equals 3x -x -5x squared?
-2
How do you prove that the line y equals x-4 is tangent to the curve of x squared plus y squared equals 8?
equation 1: y = x-4 => y2 = x2-8x+16 when both sides are squared equation 2: x2+y2 = 8 Substitute equation 1 into equation 2: x2+x2-8x+16 = 8 => 2x2-8x+8 = 0 If the discriminant of the above quadratic equation is zero then this is proof that the line is tangent to the curve: The discriminant: b2-4ac = (-8)2-4*2*8 = 0 Therefore the discriminant is equal to zero thus proving that the line is tangent to the curve.
What is the gradient of the tangent to the curve at x equals 2 if Y equals x2?
Gradient to the curve at any point is the derivative of y = x2 So the gradient is d/dx of x2 = 2x. When x = 2, 2x = 4 so the gradient of the tangent at x = 2 is 4.
How do you find the slope of an Indifference curve?
You find the tangent to the curve at the point of interest and then find the slope of the tangent.
