Parametric Equations
The parametric equations of a quadratic polynomial, parabola
The parametric equations of the parabola, whose axis of symmetry is parallel to the y-axis
The parametric equations of the parabola, whose axis of symmetry is parallel to the x-axis
The parametric equations of a quadratic polynomial, parabola
The parametric equations of the parabola, whose axis of symmetry is parallel to the y-axis
The quadratic polynomial  y = a2x2 + a1x + a0   or   y - y0a2(x - x0)2,   V(x0, y0)
are the coordinates of translations of the source quadratic  y = a2x2,  can be transformed to the parametric form by substituting  x - x0 = t.
 Therefore,
 x = t + x0 y = a2t2 + y0
 are the parametric equations of the quadratic polynomial.
Example:  Given are the parametric equations,  x = t + 1  and   y = - t2 + 4, draw the graph of the curve.
Solution:  The equation  x = t + 1 solve for t and plug into  y = - t2 + 4, thus
t = x - 1  =>   y = - t2 + 4,     y = - (x - 1)2 + 4
i.e.,  y - 4 = - (x - 1)2  or  y = - x2 + 2x + 3  translated parabola with the vertex V(x0, y0), so V(1, 4).
The parametric equations of the parabola, whose axis of symmetry is parallel to the x-axis
The quadratic expression  y2 = 2px, where p is the distance between focus and directrix, represents the source or the vertex form of the conic section called parabola with the vertex at the origin whose axis of symmetry coincide with the x-axis.
If we rewrite the above equation into  x = ay2  then,
 x = ay2 + by + c   or   x - x0 = a(y - y0)2
represents the translation of the parabola in the direction of the coordinate axes by x0 and y0 i.e., V(x0, y0).
Thus, the parabola, whose axis of symmetry is parallel to the x-axis, can be transformed to parametric form by substituting  y - y0 = t  into the above equation.
 Therefore,
 x = at2 + x0 y = t + y0
 are the parametric equations of the parabola.
Example:  Given is the parabola  x = - y2 + 2y + 3, write its parametric equations and draw the graph.
Solution:  Rewrite given equation by calculating the coordinates of translations x0 and y0 or using completing the square method, we get  x - 4 = - (y - 1)2,  where the vertex V(x0, y0), so V(4, 1).
By substituting  y - 1 = t  obtained are

 x = - t2 + 4 y = t + 1
 the parametric equations of the parabola.
Pre-calculus contents C