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For a parametric equation of a parabola in general position see § As the affine image of the unit parabola. The implicit equation of a parabola is defined by an irreducible polynomial of degree two: + + + + + =, such that =, or, equivalently, such that + + is the square of a linear polynomial.
A family of conic sections of varying eccentricity share a focus point and directrix line, including an ellipse (red, e = 1/2), a parabola (green, e = 1), and a hyperbola (blue, e = 2). The conic of eccentricity 0 in this figure is an infinitesimal circle centered at the focus, and the conic of eccentricity ∞ is an infinitesimally separated ...
Define b by the equations c 2 = a 2 − b 2 for an ellipse and c 2 = a 2 + b 2 for a hyperbola. For a circle, c = 0 so a 2 = b 2, with radius r = a = b. For the parabola, the standard form has the focus on the x-axis at the point (a, 0) and the directrix the line with equation x = −a. In standard form the parabola will always pass through the ...
Regardless of the format, the graph of a univariate quadratic function () = + + is a parabola (as shown at the right). Equivalently, this is the graph of the bivariate quadratic equation = + +. If a > 0, the parabola opens upwards. If a < 0, the parabola opens downwards.
This is the equation of a parabola, so the path is parabolic. The axis of the parabola is vertical. If the projectile's position (x,y) and launch angle (θ or α) are known, the initial velocity can be found solving for v 0 in the afore-mentioned parabolic equation:
Parametric equations are commonly used in kinematics, where the trajectory of an object is represented by equations depending on time as the parameter. Because of this application, a single parameter is often labeled t ; however, parameters can represent other physical quantities (such as geometric variables) or can be selected arbitrarily for ...
In contrast, the graph of the function f(x) + k = x 2 + k is a parabola shifted upward by k whose vertex is at (0, k), as shown in the center figure. Combining both horizontal and vertical shifts yields f(x − h) + k = (x − h) 2 + k is a parabola shifted to the right by h and upward by k whose vertex is at (h, k), as shown in the bottom figure.
In 1659 van Heuraet published a construction showing that the problem of determining arc length could be transformed into the problem of determining the area under a curve (i.e., an integral). As an example of his method, he determined the arc length of a semicubical parabola, which required finding the area under a parabola. [9]