Search results
Results from the WOW.Com Content Network
Parabola (magenta) and line (lower light blue) including a chord (blue). The area enclosed between them is in pink. The chord itself ends at the points where the line intersects the parabola. The area enclosed between a parabola and a chord (see diagram) is two-thirds of the area of a parallelogram that surrounds it.
A parabolic segment is the region bounded by a parabola and line. To find the area of a parabolic segment, Archimedes considers a certain inscribed triangle. The base of this triangle is the given chord of the parabola, and the third vertex is the point on the parabola such that the tangent to the parabola at that point is parallel to the chord.
Parabolic area: The area between the curve = and the line = Semiparabolic area. The area between the curve = and the axis, from = to = Parabolic spandrel: The ...
The area bounded by the intersection of a line and a parabola is 4/3 that of the triangle having the same base and height (the quadrature of the parabola); The area of an ellipse is proportional to a rectangle having sides equal to its major and minor axes;
A parabola may also be defined in terms of its focus and latus rectum line (parallel to the directrix and passing through the focus): it is the locus of points whose distance to the focus plus or minus the distance to the line is equal to 2a; plus if the point is between the directrix and the latus rectum, minus otherwise.
In this position, the hyperbolic paraboloid opens downward along the x-axis and upward along the y-axis (that is, the parabola in the plane x = 0 opens upward and the parabola in the plane y = 0 opens downward). Any paraboloid (elliptic or hyperbolic) is a translation surface, as it can be generated by a moving parabola directed by a second ...
The area of the surface of a sphere is equal to four times the area of the circle formed by a great circle of this sphere. The area of a segment of a parabola determined by a straight line cutting it is 4/3 the area of a triangle inscribed in this segment. For the proofs of these results, Archimedes used the method of exhaustion attributed to ...
The total torque exerted by the triangle is its area, 1/2, times the distance 2/3 of its center of mass from the fulcrum at =. This torque of 1/3 balances the parabola, which is at a distance 1 from the fulcrum. Hence, the area of the parabola must be 1/3 to give it the opposite torque.