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Q: What is tangent when x equals 4 and y equals -1?

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(2, -2)

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.

Evaluate x/y when x = 4 and y = 4. x/y = 4/4 = 1/1 = 1

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

x=4 y=1 Therefore, 6x+5y=? Just substitute, and get 6*4+5*1=29

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Cotangent 32 equals tangent 0.031

0.5

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.

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

(2, -2)

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.

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

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.

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

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)

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

x = 15.01, approx.