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The parabola opens upward. It is shown elsewhere in this article that the equation of the parabola is 4fy = x 2, where f is the focal length. At the positive x end of the chord, x = c / 2 and y = d. Since this point is on the parabola, these coordinates must satisfy the equation above.
If the parabola is tangent to the x-axis, there is a double root, which is the x-coordinate of the contact point between the graph and parabola. If the parabola does not intersect the x-axis, there are two complex conjugate roots. Although these roots cannot be visualized on the graph, their real and imaginary parts can be. [17]
A similar but more complicated method works for cubic equations, which have three resolvents and a quadratic equation (the "resolving polynomial") relating and , which one can solve by the quadratic equation, and similarly for a quartic equation (degree 4), whose resolving polynomial is a cubic, which can in turn be solved. [14]
From the point of view of projective geometry, an elliptic paraboloid is an ellipsoid that is tangent to the plane at infinity. Plane sections. The plane sections of an elliptic paraboloid can be: a parabola, if the plane is parallel to the axis, a point, if the plane is a tangent plane. an ellipse or empty, otherwise.
The graph of a real single-variable quadratic function is a parabola. If a quadratic function is equated with zero, then the result is a quadratic equation. The solutions of a quadratic equation are the zeros (or roots) of the corresponding quadratic function, of which there can be two, one, or zero. The solutions are described by the quadratic ...
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 ...
For solving the cubic equation x 3 + m 2 x = n where n > 0, Omar Khayyám constructed the parabola y = x 2 /m, the circle that has as a diameter the line segment [0, n/m 2] on the positive x-axis, and a vertical line through the point where the circle and the parabola intersect above the x-axis.
The line with equation ax + by + c = 0 has slope -a/b, so any line perpendicular to it will have slope b/a (the negative reciprocal). Let (m, n) be the point of intersection of the line ax + by + c = 0 and the line perpendicular to it which passes through the point (x 0, y 0). The line through these two points is perpendicular to the original ...