show, using algebra, that the formula for a parabola is a special case of the formula for an ellipse

1 Answer | Add Yours

mathsworkmusic's profile pic

mathsworkmusic | (Level 2) Educator

Posted on

The parametric form of an ellipse (centred on the origin) is given by

`x = a sin t`, `y = b cos t`  where `t in [0,2pi]` and `a` and `b` are constants.

The parametric form of a parabola (centred on the origin also) is given by

`x = c u^2`, `y = 2cu`  where `u in [-infty,+infty]` and `c` and `d` are constants.

Equating the `x` variables and `y` variables we get

1) ` cu^2 = asint`

2) `2cu = b cos t`

Dividing 1) by 2) gives `u = ((2a)/b) tan t`

because  `(sint)/(cost) = tan t`

The range of `t` is now restricted to `t in [-pi/2, pi/2]` because the tangent is undefined outside this range. This gives `u in [-infty,+infty]` (and the values of `a` and `b` are irrelevant as they simply determine the "speed" with which we get to the ends of the parabola - which we never do!). Since the range of `t` is restricted to half it's usual range (for a full ellipse), this is why the parabola looks like half of an ellipse.

Written in parametric form, an ellipse and parabola are equivalent when we write

u = tan(t)   where t is the parameter of the ellipse and u the parameter of the parabola. The range of t is `[-pi/2,pi/2]` and the range of u is `[-infty,+infty]`

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question