Standard Tangent Equation Of Parabola??

With the equation of the parabola in that form, take the derivative of both sides, with respect to x. You get 2y*dy/dx = 2p, which you can rearrange to get dy/dx = p/y, the slope of the tangent line. Since you've been given a point (X0, Y0), you can use the point-slope form of a line, y - Y0 = m(x - X0), where m is the slope you calculated as p/y, and you use the coordinates of your point (X0, Y0).

y^2=2px is the standard equation of a parabola whose vertex is at the origin and whose axis of symmetry is the x-axis. So it is a special case. If the vertex is on the origin and the axis of symmetry is the y-axis then the equation is x^2=2py.

For the case y^2 = 2px, you can differentiate implicitly to get

dy/dx = p/y =slope

So y-y0=(p/yo)(x-x0) is the required equation.

This form of the equation is helpful in finding the focus and the equation of the directrix as the equation of the directrix is

x=(-1/2)p and the focus is at (p/2,0).

Using the equation ax^2 +bx +c for a parabola is sometimes easier to use since it does not require the vertex to be at the origin. Here the axis of symmetry is x= -b/2a and the y- coordinate of the vertex is found by plugging this value into the parabola's equation. The slope is 2ax + b and so you can find the tangent at any point (x0,y0) by using the point-slope equation of a line.

The equation of the tangent through (x0, y0) to the parabola y^2 = 2px is

2px - 2y0 y + (y0)^2 = 0

or equivalently

px - y0 y = px0 - (y0)^2.

(That's because the eqn of tangent through (x0, y0) to the curve F(x, y) = c is: gradF(x0, y0).(x-x0, y-y0) = 0.)

f'(x)= 4x-7 f'(-2) = -15 y= -15x+b as far as I bear in suggestions this section is physically powerful... i'm no longer sur a thank you to define B tho, what incorporates my head is: 26= -15*(-2)+b 26 + 15*(-2) = b -4 = b y= -15x - 4 i'm no longer sur for the final section tho...