

Parametric Equations 
The parametric equations of a
quadratic
polynomial,
parabola 
The
parametric equations of the parabola, whose axis of symmetry is parallel
to the yaxis

The parametric equations of the
parabola, whose axis of symmetry is
parallel to the xaxis 






The parametric equations of a
quadratic
polynomial,
parabola 
The parametric equations of the parabola
whose axis of symmetry is parallel
to the yaxis

The quadratic polynomial y
=
a_{2}x^{2}
+ a_{1}x + a_{0}
or y

y_{0}
= a_{2}(x
 x_{0})^{2},
V(x_{0},
y_{0}) 

are
the coordinates of translations of the source quadratic y
=
a_{2}x^{2},
can be transformed to the parametric 
form
by
substituting x
 x_{0}
= t. 

x
=
t + x_{0} 
y
=
a_{2}t^{2} + y_{0} 


are the
parametric equations of the
quadratic polynomial. 



Example: Given
are the parametric equations, x
=
t + 1
and y
=
 t^{2} + 4,
draw the graph of the curve. 
Solution:
The equation x
=
t + 1
solve for t
and plug into y
=
 t^{2} + 4,
thus 
t
=
x
 1
=> y
=
 t^{2} + 4,
y
=
 (x
 1)^{2} + 4 
i.e., y

4
=
 (x
 1)^{2}
or y
=
 x^{2} + 2x + 3 translated parabola with the vertex V(x_{0},
y_{0}),
so V(1,
4). 


The parametric equations of the parabola, whose axis of symmetry is
parallel to the xaxis 
The
quadratic expression y^{2}
=
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 xaxis. 
If
we rewrite the above equation into x
= ay^{2} then, 

x
= ay^{2} + by + c^{ }
or
x
 x_{0}
= a(y
 y_{0})^{2} 



represents
the translation of the parabola in the direction of the coordinate axes
by x_{0}
and y_{0}
i.e.,
V(x_{0},
y_{0}). 
Thus,
the parabola, whose axis of symmetry is parallel to the xaxis,
can be transformed to parametric form by
substituting y
 y_{0}
= t
into the above equation. 

x
=
at^{2}
+ x_{0} 
y
= t
+ y_{0} 


are the parametric equations
of the parabola. 



Example: Given
is the parabola x
=
 y^{2} + 2y + 3,
write its parametric equations and draw the graph. 
Solution:
Rewrite given equation by calculating the coordinates of
translations x_{0}
and y_{0}
or using completing the
square method, we get x

4
=
 (y
 1)^{2},
where the vertex V(x_{0},
y_{0}),
so V(4,
1). 
By
substituting y
 1
= t
obtained are 

x
=

t^{2}
+ 4 
y
= t
+ 1 


the parametric equations of the parabola. 












Functions
contents B




Copyright
© 2004  2020, Nabla Ltd. All rights reserved. 