The line 2x + 3y = 13 cuts the curve 5y^2 = 4x^2 - 2xy + 41 at the points C and D. Find the gradient of the line CD.

Answer 1

To find the points of intersection between the line and the curve, we can first rewrite the equation of the curve in terms of x:
5y^2 = 4x^2 - 2xy + 41
5y^2 + 2xy - 4x^2 = 41
y(5y + 2x) - 4x^2 = 41
y = (4x^2 - 41) / (5y + 2x)

Now we substitute y = (13 - 2x) / 3 into this equation:
(13 - 2x) = (4x^2 - 41) / (5(13 - 2x) + 2x)
13 - 2x = (4x^2 - 41) / (65 - 10x + 2x)
13 - 2x = (4x^2 - 41) / (65 - 8x)
13(65 - 8x) - 2x(65 - 8x) = 4x^2 - 41
845 - 104x - 130 + 16x = 4x^2 - 41
715 - 88x = 4x^2
4x^2 + 88x - 715 = 0
x^2 + 22x - 178.75 = 0

Now we can solve for x using the quadratic formula:
x = (-22 ± √(22^2 - 4(1)(-178.75))) / 2(1)
x = (-22 ± √(484 + 715)) / 2
x = (-22 ± √1199) / 2

This gives us two x-coordinates for the points C and D. By substituting these values back into the equation of the line 2x + 3y = 13, we can find the corresponding y-coordinates.

The equation x = (-22 ± √1199) / 2 provides the x-coordinates of the points C and D. By substituting these x-values into the equation 2x + 3y = 13, one can determine the respective y-coordinates for points C and D.

Once the coordinates of points C and D are determined, the gradient of the line CD can be calculated using the formula (y2 - y1) / (x2 - x1) for the two points.