The equations for any conic section (which includes both parabolas and circles) can be written in the following form: Ax^2+Bxy+Cy^2+Dx+Ey+F=0 Some terms might be missing, in which case their coefficient is 0. The way to figure out if the equation is a parabola, circle, ellipse, or hyperbola is to look at the value of B^2-4AC: If it's negative, the graph is an ellipse (of which a circle is a special case). If it's 0, the graph is a parabola. If it's positive, the graph is a hyperbola. The special case of a circle happens when B is 0 -- there is no "xy" term -- and A=C.
A gem of a question.The equation of the parabola is:y = x2And the general equation of a circle that intersects the parabola at the origin, and points upwards in the y direction, is:x2 + (y-r)2 = r2That is: x2 + y2= 2yrNow, if the circle could have been 'dropped' into the parabola from above, then it must intersect with the parabola exclusively at the origin. If, on the other hand, the circle intersects the parabola elsewhere, it must be partially below the curve of the parabola, and hence could not nest within it.Note that the whole situation is symmetrical about the y axis. If an intersection occurs on one side of the axis then it occurs on the other. It is sufficient, therefore, to consider the possible intersections in the first quadrant.The equation of the circle in the first quadrant is:x=sqrt(2yr - y2)And the parabola:x=sqrt(y)Solving these simultaneously for any intersections we get:sqrt(2yr - y2) = sqrt(y)Therefore: 2yr - y2 = y and y(2r - y - 1) = 0Clearly one solution to this is y = 0. The other root of the equation is given by solving2r - y - 1 = 0y = 2r - 1In order for there to be no intersections other than the origin in the first quadrant, this root must be less than or equal to zero.0 > 2r - 1r < 0.5Hence the radius of the biggest possible circle that will nest in y=x2 is 0.5.(In general, by the way, for y = Ax2, the largest radius is 1/2A. This result can also be obtained by considering the second differential of the two functions.)The smallest radius is 1/2A , at the bottom. So in this case (A=1) and therefore the smallest radius is 1/2 and any circle with a greater radius won't fit inside at the bottom.
i tested this and yes i did!Yes, even looking at red surfaces..
I'm assuming you are looking for the name of the conic section produced by this type of intersection? If a right circular cone is intersected by a plane parallel to one edge of the cone, the resulting curve of intersection would be a parabola. If the intersecting plane was parallel to the base, it would be a circle. If the intersecting plane was at any angle between being parallel to the base and being parallel to an edge, it would produce an ellipse or part of an ellipse (depending on whether the intersection was completely within the cone).
This is a fairly straightforward trigonometry problem. I'll assume you were give tan(θ)=(-1) for 0o < θ < 360o in this situation. I strongly suggest you familiarize yourself with the unit circle. In this case, we are looking for a point on the circle where the slope between (0,0) and the point at θ is (-1). This occurs at 135o and 315o. Short Answer: θ = {135o, 315o}
A cone has a verticy and a big circle face a cylinder has no verticys just 2 circle faces and a curved face in the middle
It has 2 pointsA2. It IS NOT a parabola. It is an arc of a circle. As many mountaineers or plane flyers know by looking down on the top of clouds and seeing a perfect circle of rainbow light surrounding their shadow.
In its standard form, the equation of a circle is a quadratic in both variables, x and y, whereas a parabola is quadratic in one (x) and liner in the other (y). A circle is a closed shape and comprises the locus of all points that are equidistant from one given point (the centre). A parabola is an open shape and comprises the locus of all points that are the same distance from a a straight line (the directrix) and a point not on that line (the focus).
its orbit is not perfect circle, but is most eccentric of all planets,having eccentricity of 0.21 eccentricity means,the amount by which its orbit varies from perfect circle. 0 means circle and 1 means parabola. so mercury's eccentricity(0.21) is between circle and parabola, in fact, more closer to circle
No, it is not. It is an arc of a circle.
Sources vary. According to some, it is part (arc) of a circle, not a parabola. According to college professors and many proven resources online, it is a Parabola.
Relationship between radius and area of a circle is nonlinear. Area = pi * radius^2, so it is like a quadratic. If you graphed radius on the horizontal, and area on the vertical, it would be a parabola (actually a half of a parabola, since you cannot have a negative radius).
Yes.
No, there is no parabola shape on the arc. The legs are parallel lines connecting to the arch which is an half-circle.
That can be an ellipse, circle, parabola, or hyperbola. It depends on the angle between the plane and the axis or surface of the cone.
They are all conic sections.
No. It can also be a circle, ellipse or hyperbola.
Linear equations or inequalities describe points x y that lie on a circle.