The known curve {x = 2T, y = 2-T, t is a parameter The known curve {x = 2T, y = 2-T, t is the intersection of the parameter and X axis, Y axis and two points of AB, C moves up on the curve xv2 = - 4Y to find the minimum ABC of the triangle

The known curve {x = 2T, y = 2-T, t is a parameter The known curve {x = 2T, y = 2-T, t is the intersection of the parameter and X axis, Y axis and two points of AB, C moves up on the curve xv2 = - 4Y to find the minimum ABC of the triangle

A (4,0), B (0,2). The parametric curve is a straight line y = - X / 2 + 2. If AB is regarded as the bottom edge of triangle ABC, the minimum area is obtained when the height is minimum. At this time, the tangent of point C is parallel to the straight line, that is, the tangent slope of point C, that is, the derivative is - 1 / 2. Because y '= - X / 2 = - 1 / 2, the distance from C to the straight line y + X / 2-2 = 0 is - (- 1 + 1 / 2-2) / radical (1 + 1 / 4) = radical 5, and ab = 2 radical 5, so the minimum area is 1 / 2 * radical 5 * 2 radical 5 = 5

When passing through point a (- 1, - 3), the slope is - 1 / 4 of the slope of the line y = 3x

The slope of the line y = 3x is 3, and its - 1 / 4 is - 3 / 4
Let the linear equation be y = - 3 / 4x + B
Because it passes through point a (- 1, - 3),
So - 3 = 3 / 4 + B
The solution is b = - 15 / 4
So the linear equation is y = - 3 / 4x-15 / 4

Solve the linear equation of P (3,2) and the slope is twice the slope of the line y = X-1

If the slope of the line y = X-1 is 1, then the equation of the line P (3,2) with the slope of 2 is given by
The slope of a straight line is known to be K in the point oblique form, and the equation of a straight line passing through the point (x0, Y0) is as follows:
y-y0=k(x-x0)
y-2=2(x-3)
y=2x-4

It is known that x ^ 2sina-y ^ 2cosa = 1 (0, a)

x²/(1/sina)-y²/(1/cosa)=1
The focus is on the y-axis
y²/(-1/cosa)-x²/(-1/sina)=1
So - 1 / cosa > 0, - 1 / Sina > 0
That's Sina