Answer:
Explanation:
The latus rectum of the parabola y^2 = 6x is a line segment passing through the focus (3,0) and perpendicular to the axis of the parabola. The length of the latus rectum is 4p, where p is the distance from the vertex to the focus, and in this case, p = 3/2. Therefore, the positive end of the latus rectum is located at the point (3, 2p) = (3, 3).
To find the equation of the tangent to the parabola at this point, we need to first find the slope of the tangent. We can do this by taking the derivative of y^2 = 6x with respect to x:
d/dx (y^2) = d/dx (6x)
2y (dy/dx) = 6
Solving for dy/dx, we get:
dy/dx = 3/y
At the point (3,3), the slope of the tangent is:
dy/dx = 3/3 = 1
Therefore, the equation of the tangent to the parabola at the positive end of the latus rectum is:
y - 3 = 1(x - 3)
or
y = x - 3
Note that we have used the point-slope form of the equation of a line, with the point (3, 3) on the line and the slope of the tangent equal to 1.