Parametric coordinates of a parabola

Question

I was recently studying parabolas when I came across a question, in which we were supposed to find the coordinates of the point of contact of the tangent $ y=1-x$ with the parabola $y^2-y+x=0$. In the question, the parametric coordinates of the point of contact were given as $$\left (\frac{1}{4}-\frac{a}{m^2},\frac{1}{2}-\frac{2a}{m}\right)$$ whereas in the same book the parametric coordinates for a parabola $$(y-k)^2=-4a(x-h)$$ are given as $$\left (h-\frac {a}{m^2},k+\frac{2a}{m}\right)$$ how is this even possible?

P.S The answer is given as $(0,1)$

By completing square, the equation of parabola can be written as $(y-\frac{1}{2})^2=4(-\frac{1}{4})(x-\frac{1}{4})$. — SarGe, Aug 28 '20 at 16:32
even if I rename m to -m both expressions will change and they will still not be similar — ErrorEliminator, Aug 28 '20 at 17:07
With $h\to {1\over4}$, $k\to {1\over2}$ and $m\to-m$ your second formula becomes the first one. — Intelligenti pauca, Aug 28 '20 at 20:59

Narasimham · Answer 1 · 2020-08-28T17:57:25.193

1

HINT:

To avoid infinite slopes put $m= \dfrac1t$ to a more conveniently familiar form.

It gives a parabola parametrization with symmetry axis parallel to x-axis

$$(x,y)=(h-at^2, k+2at)$$

and you can see where the $(h,k)$ displacements occur as well as slope

$$=- \dfrac{1}{t}=-m$$

at tangent point $(0,1)$.

edited Aug 28 '20 at 17:57

answered Aug 28 '20 at 16:58

Narasimham

40,495

score 0 · Answer 2 · answered Aug 28 '20 at 16:42

0

To find the coordinates of the contact point we solve the system $$\left\{ \begin{array}{l} x = 1 - y\\ {y^2} - y + x = 0 \end{array} \right.$$ $y^2-y+1-y=0$

$y^2-2y+1=0$

$(y-1)^2=0$

$y=1$ and $x= 0$

answered Aug 28 '20 at 16:42

Raffaele

26,371

I know this method and have also solved using it, but my confusion is in the parametric coordinates – ErrorEliminator Aug 28 '20 at 17:06

Parametric coordinates of a parabola

2 Answers2