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It is 8*sqrt(2)/3 = 3.7712 approx.

Q: Find the area in the first quadrant bounded by the curve y squared equals 4x the x axis and the line x equals 2?

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k = 0.1

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)

The line x-y = 2 intersects with the curve x^2 -4y^2 = 5 at (2.5, 1/3) and (3, 1) and by using the distance formula its length is 5/6

They work out as: (-3, 1) and (2, -14)

2

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(2, -2)

It is (-0.3, 0.1)

k = 0.1

If you mean the coordinates of the line x-y = 2 that intersects the curve of x2-4y2 = 5 Then the coordinates work out as: (3, 1) and (7/3, 1/3)

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)

The line x-y = 2 intersects with the curve x^2 -4y^2 = 5 at (2.5, 1/3) and (3, 1) and by using the distance formula its length is 5/6

They work out as: (-3, 1) and (2, -14)

2

If you mean a line of y = 2x+5/4 and a curve of y^2 = 10x Then it works out that the line touches the curve at: (5/8, 5/2)

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

-2

A quadratic equation (t=s2+3). This kind of line will result in a parabola like curve.