Standard Tangent Equation Of Parabola??

Question

Given parabola y^2=2px (standard parabola equation), and a point on parabola P(X0, Y0)
Ask for the equation of tangent line passing through P(X0, Y0)

I'v worked out the equation but it looks complex, unlike tangent equation of ellipse's and hyperbola's, which are very standardised and similar to that of the conics.
(say:tangent equation of ellipse:
X0X/a^2+Y0Y/b^2=1 )

So I wanna make sure whether I had it right? or tangent equation of parabola doesn't look as standardised as those of other conics?

Answer 1

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).

Answer 2

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.

Answer 3

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.)

Answer 4

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...