How to derive formula for parabola?
if the sum of the distances from the parabola to the focus and directrix are constant D and the angle between the two distance lines is 90 degrees when the parabolic point has the same y coordinate as the focus the distance front focus to parabola and from directrix to parabola should be equal to each other and equal to D/2 right angles with adjacent sides equal are two edges of a square so you should be able to construct that point by drawing a square with one corner on the focus and one side sitting on the directrix, or it's algebraic/arithmetic equivalent so you get an exact expression for it. That would get you a 2nd point on the parabola. I don't feel like checking this with a formal proof right now.
What is the formula for a parabola?
The general formula is y - k = a(x - h)^2 where (h,k) is the vertex. I've seen other books write it as y = a(x - h)^2 + k, or y - k = 4p(x - h)^2, but it's the same idea. Also a quadratic in general, y = ax^2 + bx + c, will give you a parabola so long as a is not 0. So that's another way of writing it. In both cases, the parabola opens upwards if a>0 and opens downwards if a<0. I see a lot of wrong answers here. Somebody said y=mx+b, which is a straight line, not a parabola. Others are saying things like "y=x^2" which gives you a parabola, but is not the general form.
the slope of a parabola like the curve of x^2 is found by a drawing a tangent line to the curve and find the instantaneous or Average rate of change which indirectly represents the slope of the parabola.the above represents graphical method for finding slopes but if it stated in equation form,you find the first derivative and find the slope at the given interval.
The volume of a paraboloid is as follows:V = 1/2 * pi * b^2 * aCLICK HERE for an on-line calculator that computes the volume of a paraboloid with the option of many different length and volume units.
To find the centre of a conic, you need to first partially differentiate the equation of the conic wrt one variable and then do the same wrt another variable and then solve both the equations simultaneously.In this case, let the equation of the parabola be represented by f(x,y)f(x,y) = y*y - 4*a*x where a is an arbitrary constant not equal to 0.Partial differentiation of f(x,y) wrt x yields -4*a which can never be equal to 0. Hence, there is no centre of parabola.
Messure 3 points, specially the one where it crosses the y-axes.It has the general equation:[math]ax^2+bx+c=y[/math]So you can make a linear system and solve for [math]a[/math], [math]b[/math] and [math]c[/math]. For example let’s say you got [math](-3, 5)[/math], [math](6, 9)[/math] and [math](0, 7)[/math] than the function would be [math]\frac{-(x^2)}{27}+\frac{5x}{9}+7[/math].
The y intercept of any parabola can be obtained by substituting the value of x=0 wich is the eqaution of y axis. FYI three cases will be possible.1.THE PARABOLA TOUCHES THE Y AXIS IN 2 POINTSHere when you substitute x=0,you will get a quadratic equation in y having 2 real values.2.THE PARABOLA TOUCHES THE Y AXIS IN 1 POINTHere you will get one and reapeating values of y.3.THE PARABOLA DOES NOT TOUCH THE Y AXIS.Here the quadratic eqation roots will be imaginary.