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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 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 evaluate and write this question in simplest form x over y for x equals 4 and y equals 4 what is the answer?
Evaluate x/y when x = 4 and y = 4. x/y = 4/4 = 1/1 = 1
What are the tangent line equations of the circle x2 plus y2 -6x plus 4y plus 5 equals 0 when the circle passes through the x axis?
Equation of circle: x^2 +y^2 -6x +4y +5 = 0 Completing the squares: (x-3)^2 +(y+2)^2 = 8 Radius of circle: square root of 8 Center of circle: (3, -2) Circle makes contact with the x axis at: (1, 0) and (5, 0) Slope of 1st tangent: 1 Slope of 2nd tangent: -1 1st tangent line equation: y = 1(x-1) => y = x-1 2nd tangent line equation: y = -1(x-5) => y = -x+5
If x equals 4 and y equals 1 then 6x plus 5y equals?
x=4 y=1 Therefore, 6x+5y=? Just substitute, and get 6*4+5*1=29
What is cotangent 32 equals tangent x?
Cotangent 32 equals tangent 0.031
Find the equation of the line tangent to the graph of y equals cos2x at x equals pi divided by 4?
0.5
What line is the tangent to y equals x squared when x equals 2?
To get the tangent, you first need to find it's slope: y = x2 ∴ dy/dx = 2x So the slope with respect to x is equal to 2x. When x = 2 then, the slope is 4. Now you need the y-coordinate that occurs when x equals 2. We know that y = x2, so when x equals 2, y equals 4. You now have a point on the tangent, as well as it's slope, allowing you to define the line: Recall the definition of a line: Δy = sΔx or: y - y1 = s(x - x1) ∴ y - 4 = 4(x - 2) ∴ y = 4 + 4x - 8 ∴ y = 4x - 4 And that is your tangent.
What is the radius equation inside the circle x squared plus y squared -8x plus 4y equals 30 that meets the tangent line y equals x plus 4 on the Cartesian plane showing work?
Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius
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 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.
What is the distance from a defined point on the x axis to the centre of circle x2 plus y2 -2x -6y plus 5 equals 0 when its tangent is at 3 4 on the Cartesian plane?
Equation of circle: x^2 +y^2 -2x -6y +5 = 0 Completing the squares: (x-1)^2 +(y-3)^2 = 5 Center of circle: (1, 3) Tangent contact point: (3, 4) Slope of radius: ((3-4)/(1-3) = 1/2 Slope of tangent line: -2 Equation of tangent line: y-4 = -2(x-3) => y = -2x+10 Equation tangent rearranged: 2x+y = 10 When y equals 0 then x = 5 or (5, 0) as a coordinate Distance from (5, 0) to (1, 3) = 5 using the distance formula
If x2 plus y4 equals 10x plus 7 then what is the slope of the tangent line at 1 -2 and what is the equation of that tangent line- and at what point are the lines tangent to the original equation?
x^2+y^4=10x+7 @ (1,-2) Rearrange to isolate y. y=(-x^2+10x+7)^(1/4) Take derivative. y'=(1/4)*((-x^2+10x+7)^(-3/4))*(-2x+10) y'=(-2x+10)/4*(-x^2+10x+7)^(3/4) Sub in x=1. y'=8/2*(16)^(3/4) y'=2/8 y'=1/4 y'=m=slope=1/4 Sub m and (1,-2) into linear equation (tangent line). m*(x-x1)=y-y2 (1/4)(x-1)=y+2 Isolate y. Equation of tangent line: y=(1/4)x-(9/4). The third question is unclear.
What are the equations of the tangent and normal at the coordinate of 6 4 of the circle x2 -4x plus y2 -6y equals 4 when plotted on the Cartesian plane?
Circle equation: x^2 -4x +y^2 -6y = 4 Completing the squares: (x-2)^2 +(y-3)^2 = 17 Center of circle: (2, 3) Point of contact: (6, 4) Slope of radius: 1/4 Slope of tangent: -4 Slope of normal: 1/4 Equation of tangent: y-4 = -4(x-6) => y = -4x+28 Equation of normal: y-4 = 1/4(x-6) => 4y-16 = x-6 => 4y = x+10
What is the point of contact when the tangent line y equals x plus 4 meets the circle x squared plus y squared -8x plus 4y equals 30?
Circle equation: x^2 +y^2 -8x +4y = 30 Completing the squares: (x-4)^2 +(y+2)^2 = 50 Centre of circle: (4,-2) Slope of radius: -1 because it's perpendicular to the tangent line Radius equation: y--2 = -1(x-4) => y = -x+2 If: y = x+4 and y = -x+2 Then: x+4 =-x+2 or 2x = -2 => x = -1 Therefore by substitution the point of contact occurs at: (-1, 3)
What is the slope of the tangent to the curve x squared plus y squared equals 26 at x 0 equals 4 in the first quadrant. Round the answer to 3 decimal places.?
Assuming you mean the tangent at x = 4 of x² + y² = 26 then: x² + y² = 26 → y² = 26 - x² → y = √(26 - x²) → slope = dy/dx = d/dx √(26 - x²) = -x/√(26 - x²) At x = 4: slope = -4/√(26 - 4²) = -1.26491106.... ≈ -1.265
What is tangent 25 equals 7 divided by x?
x = 15.01, approx.
