Find the slope of the line tangent to the curve (2x+1)2−(3y−1)2=24 at the point (3,2)
To find the slope of the line tangent to the curve at the point (3, 2), we need to find the derivative of the curve and then evaluate it at the given point.
First, let's simplify the equation of the curve:
(2x + 1)^2 - (3y - 1)^2 = 24
4x^2 + 4x + 1 - 9y^2 + 6y - 1 = 24
4x^2 + 4x - 9y^2 + 6y = 24
Next, we find the derivative of this equation with respect to x:
d/dx(4x^2 + 4x - 9y^2 + 6y) = d/dx(24)
8x + 4 - 18y(dy/dx) + 6(dy/dx) = 0
Now we substitute the point (3, 2) in this equation:
8(3) + 4 - 18(2)(dy/dx) + 6(dy/dx) = 0
24 + 4 - 36(dy/dx) + 6(dy/dx) = 0
28 - 30(dy/dx) = 0
30(dy/dx) = 28
(dy/dx) = 28/30
= 14/15
Therefore, the slope of the line tangent to the curve at the point (3, 2) is 14/15.