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The equation of a parabola
(higher level courseware)
(higher level courseware)
DEFINITION: A parabola is the set of points in the plane equidistant from a straight line \latex{ d } of the plane and a point \latex{ F } not lying on \latex{ d } (Figure 57).
\latex{F} is the focus (focal point), \latex{d} is the directrix of the parabola. The parabola is axially symmetric about the straight line \latex{t} passing through \latex{F} and perpendicular to \latex{d}. The straight line \latex{t} is the axis of the parabola. The midpoint \latex{T} of the line segment of the axis between \latex{F} and d lies on the parabola. The point \latex{T} is the vertex of the parabola. The distance of \latex{F} and \latex{d} is the focal parameter of the parabola, denoted by \latex{p(p\gt0)}.
In the Cartesian coordinate system the parabola defined by \latex{F} and \latex{d} can be located in many different ways depending on their positions. Now we are going to give the equation of the parabola in a few special cases, i.e. an equation in two variables that is satisfied by the coordinates of the points of the parabola, but it is not satisfied by the coordinates of any other point.
- Let the axis of the parabola with the focal parameter \latex{p} be the \latex{ y }-axis, let its vertex be the origin, and let its focus lie on the positive half of the \latex{ y }-axis. Then \latex{F\left\lgroup0;\frac{p}{2} \right\rgroup } , the equation of the directrix is \latex{y=-\frac{p}{2}} (Figure 58).
If \latex{P(x; y)} is an arbitrary point of the parabola, then based on the definition
\latex{y+\frac{p}{2}=\sqrt{x^2+\left\lgroup y-\frac{p}{2} \right\rgroup^2 } }.
By squaring it
\latex{y^2+py+\left\lgroup\frac{p}{2} \right\rgroup^2=x^2+y^2-py+ \left\lgroup\frac{p}{2} \right\rgroup^2}.
\latex{y^2+py+\left\lgroup\frac{p}{2} \right\rgroup^2=x^2+y^2-py+ \left\lgroup\frac{p}{2} \right\rgroup^2}.
\latex{y=-\frac{p}{2}}
\latex{P(x;y)}
\latex{\frac{p}{2}}
\latex{F\left\lgroup0;\frac{p}{2} \right\rgroup }
\latex{ y }
\latex{ 0 }
\latex{ x }
\latex{ y }
Figure 58
By cancelling the terms appearing on both sides and by rearranging it so that one side is equal to \latex{ y } we obtain the following:
\latex{y=\frac{1}{2p}\times x^2 }.
This equation is the equation of the parabola with the focal parameter \latex{ p } and with the focus \latex{F\left\lgroup0;\frac{p}{2} \right\rgroup }. This equation is usually called the vertex form of the equation of the parabola.
- Similarly to the previous one we can find that
- the equation of the parabola with the focal parameter \latex{ p } and with the focus \latex{F\left\lgroup0;-\frac{p}{2} \right\rgroup }(Figure 59) is
\latex{y=-\frac{1}{2p}\times x^2 }.
\latex{y=\frac{p}{2}}
\latex{F\left\lgroup0;-\frac{p}{2} \right\rgroup }
\latex{P(x;y)}
\latex{ y }
\latex{ 0 }
\latex{ x }
Figure 59
- the equation of the parabola with the focal parameter \latex{ p } and with the focus \latex{F\left\lgroup \frac{p}{2};0 \right\rgroup }(Figure 60) is
\latex{x=\frac{1}{2p}\times y^2 }.
- the equation of the parabola with the focal parameter \latex{ p } and with the focus \latex{F\left\lgroup -\frac{p}{2};0 \right\rgroup } (Figure 61) is
\latex{x=-\frac{p}{2}}
\latex{F\left\lgroup \frac{p}{2};0 \right\rgroup }
\latex{P(x;y)}
\latex{ y }
\latex{ 0 }
\latex{ x }
Figure 60
\latex{x=-\frac{1}{2p}\times y^2 }.
- Let the vertex of the parabola with the focal parameter \latex{ p }, with the axis parallel with the \latex{ y }-axis and “open upwards” be the point \latex{T(u; v)}. Then the focus of the parabola is \latex{F\left\lgroup u;v+\frac{p}{2} \right\rgroup}, the equation of its directrix is \latex{y=v-\frac{p}{2} } (Figure 62).
For an arbitrary point \latex{P(x; y)} of the parabola according to the definition
\latex{y-\left\lgroup v-\frac{p}{2} \right\rgroup =\sqrt{(x-u)^2+\left\lgroup y-\left\lgroup v+\frac{p}{2} \right\rgroup \right\rgroup^2 }}.
\latex{y-\left\lgroup v-\frac{p}{2} \right\rgroup =\sqrt{(x-u)^2+\left\lgroup y-\left\lgroup v+\frac{p}{2} \right\rgroup \right\rgroup^2 }}.
This implies that
\latex{y^2-2y\times \left\lgroup v-\frac{p}{2} \right\rgroup+\left\lgroup v-\frac{p}{2} \right\rgroup^2=(x-u)^2+y^2+2y\times \left\lgroup v+\frac{p}{2} \right\rgroup+\left\lgroup v+\frac{p}{2} \right\rgroup^2.}
\latex{y^2-2y\times \left\lgroup v-\frac{p}{2} \right\rgroup+\left\lgroup v-\frac{p}{2} \right\rgroup^2=(x-u)^2+y^2+2y\times \left\lgroup v+\frac{p}{2} \right\rgroup+\left\lgroup v+\frac{p}{2} \right\rgroup^2.}
\latex{x=\frac{p}{2}}
\latex{F\left\lgroup -\frac{p}{2};0 \right\rgroup }
\latex{P(x;y)}
\latex{ y }
\latex{ 0 }
\latex{ x }
Figure 61
After cancelling the common terms and after the addition
\latex{2\times yp-2\times vp=(x-u)^2,}
which implies
\latex{y-v=\frac{1}{2p}\times (x-u)^2 }.
The equation obtained is the equation of the parabola with the focal parameter \latex{ p }, with the vertex \latex{T(u; v)}, with the axis parallel with the \latex{ y }-axis and “open upwards”.
\latex{y=v-\frac{p}{2}}
\latex{F\left\lgroup u;v- \frac{p}{2} \right\rgroup }
\latex{P(x;y)}
\latex{T(u;v)}
\latex{ y }
\latex{ 0 }
\latex{ x }
Figure 62
Notes:
- This parabola is obtained by translating the parabola with the equation \latex{y=\frac{1}{2p}\times x^2 } by the vector \latex{\vec{r} (u;v)} (Figure 63).
- In Figure 64 the graphs and the equations of the other parabolas with the vertex \latex{T(u; v)} with the focal parameter \latex{ p } and with the axis parallel with one of the coordinate axes can be seen.
It can be proven that the equation of a parabola is always a quadratic equation in two variables, no matter where it is located in the coordinate system.
\latex{T(u;v)}
\latex{T(u;v)}
\latex{T(u;v)}
\latex{y-v=-\frac{1}{2p}\times (x-u)^2 }
\latex{x-u=\frac{1}{2p}\times (y-v)^2 }
\latex{x-u=-\frac{1}{2p}\times (y-v)^2 }
\latex{ y }
\latex{ 0 }
\latex{ x }
\latex{ 0 }
\latex{ x }
\latex{ y }
\latex{ y }
\latex{ x }
\latex{ 0 }
Figure 64
\latex{T(u;v)}
\latex{\vec{r}}
\latex{ 0 }
\latex{ x }
\latex{ y }
Figure 63
Example 1
A straight line and a point not lying on the straight line are given in the plane. Let us construct a few points of the parabola whose directrix is the given straight line and whose focus is the given point.
Solution
The focal parameter of the parabola is the distance \latex{ p } between the given point and the given straight line. The midpoint of the perpendicular line segment to the directrix from the focus is the vertex \latex{ T } of the parabola. Two points of the parabola different from \latex{ T } can be constructed as follows (Figure 65).
- We take a line segment longer than \latex{\frac{p}{2}} , with a length of \latex{ s }.
- We construct a circle with a radius of \latex{ s } with the given point (focus) as its centre.
- We intersect this circle with a straight line that is parallel with the directrix and is at a distance of \latex{ s } from it in the half-plane containing the focus. The two intersection points obtained are two points of the parabola symmetric about the axis.
\latex{ s }
\latex{ s }
\latex{ F }
\latex{ T }
\latex{ s }
\latex{ d }
\latex{ P_1 }
\latex{ P_2 }
Figure 65
Finitely many points of the parabola can be constructed with this process.
Example 2
Let us determine the vertex, the focal parameter, the focus and the equation of the directrix of the parabola with the equation \latex{y = x^2 + 2x + 2.}
Solution
By completing the square on the right-hand-side of the given equation
\latex{y = (x + 1)^2 +1}, which implies \latex{y = (x + 1)^2 +1}.
\latex{y = (x + 1)^2 +1}, which implies \latex{y = (x + 1)^2 +1}.
By comparing it to the corresponding general form of the equation the following are obtained: the vertex is \latex{T(-1;1)}, the focal parameter is \latex{p=\frac{1}{2}}, the focus is \latex{F\left\lgroup-1;\frac{5}{4} \right\rgroup } and the equation of the directrix is \latex{y=\frac{3}{4}} (Figure 66).
\latex{y=(x+1)^2+1}
\latex{y=\frac{3}{4}}
\latex{ T }
\latex{ y }
\latex{ x }
\latex{ 0 }
\latex{ F }
Figure 66
Example 3
Let us set up the equation of the parabola with the focal parameter \latex{ 2 } whose axis is parallel with the y-axis and its focus is \latex{F(–3; 5)}.
Solution
Two parabolas meet the conditions, the vertex of one of them is \latex{T_1(–3; 4)} and the vertex of the other one is \latex{T_2(–3; 6)} (Figure 67).
The equation of the parabola “open upwards” is
\latex{y-4=\frac{1}{4}\times (x+3)^2},
the equation of the parabola “open downwards” is
\latex{y-6=-\frac{1}{4}\times (x+3)^2}.
\latex{F }
\latex{ y }
\latex{ x }
\latex{1 }
\latex{0 }
\latex{-1 }
\latex{-3 }
\latex{1 }
\latex{T_1 }
\latex{T_2 }
Figure 67
Example 4
Where does the straight line with a slope of \latex{ 1 } passing through the focus of the parabola with the equation \latex{8y = (x + 1)^2} intersect the parabola?
Solution
The equation of the parabola shows that its axis is parallel with the \latex{ y }-axis, it is “open upwards”, its vertex is \latex{T(–1; 0)}, and its focal parameter is \latex{ 4 } (Figure 68).
It is easily obtained from the data that the focus of the parabola is \latex{F(–1; 2)}. Based on the gradient form the equation of the straight line with a slope of \latex{ 1 } and passing through the point \latex{ F } is:
\latex{y - 2 = x + 1}, or in another form \latex{y = x + 3.}
\latex{F(-1;2)}
\latex{y=\frac{1}{8}\cdot (x+1)^2 }
\latex{y=x+3}
\latex{ M_1 }
\latex{ y }
\latex{ x }
\latex{ 1 }
\latex{ -1 }
\latex{ M_2 }
Figure 68
The coordinates of the common points of the straight line and the parabola are the solutions of the simultaneous equations
\latex{\begin{rcases}\begin{align*}8y=(x+1)^2\\ y=x+3\end{align*}\end{rcases}}
consisting of the equations of the parabola and the straight line.
By substituting the second equation into the first one
\latex{8\times(x+3)=(x+1)^2},
and after rearranging
\latex{x^2 - 6x - 23 = 0.}
Its solutions are:
\latex{x_1=3+4\times \sqrt{2}},
\latex{x_2=3-4\times \sqrt{2}}.
\latex{x_2=3-4\times \sqrt{2}}.
By substituting these the corresponding \latex{ y }-values are obtained, and thus the two intersection points are:
\latex{M_1(3+4\times \sqrt{2};6+4\times\sqrt{2} )},
\latex{M_2(3-4\times \sqrt{2};6-4\times\sqrt{2} )}.
\latex{M_2(3-4\times \sqrt{2};6-4\times\sqrt{2} )}.
In example \latex{ 4 } the parabola and the straight line have two points in common. It is also true in general that a parabola and a straight line can have at most two points in common, as the system of equations in two variables consisting of their equations is quadratic, thus it can have at most two distinct ordered real number pairs as solution.
Example 5
Let us set up the equations of the straight lines passing through the point \latex{P(5; 8)} that have exactly \latex{ 1 } point in common with the parabola with the vertex \latex{T(5; 5)}, with the focal parameter of \latex{ 2 }, with the axis parallel with the \latex{ y }-axis and “open downwards”.
Solution
The conditions imply that the equation of the parabola is
\latex{y-5=-\frac{1}{4} \times (x-5)^2}.
As the point \latex{ P } lies on the axis with the equation \latex{ x = 5 } of the parabola, and the axis has a single point in common with the parabola (the vertex \latex{ T }), one of the solutions of the example is the axis of the parabola (Figure 69).
\latex{P(5;8)}
\latex{T(5;5)}
\latex{x=5}
\latex{ e_2 }
\latex{ y }
\latex{ 8 }
\latex{ 1 }
\latex{ 0 }
\latex{ 1 }
\latex{ x }
\latex{ e_1 }
Figure 69
The other suitable straight lines are not parallel with the \latex{ y }-axis, therefore their equations have the form
\latex{y-8=m\times(x-5)},
where \latex{ m } is the corresponding slope.
Such an equation has exactly one point in common with the parabola if the simultaneous equations
\latex{\begin{rcases}\begin{align*}y-5=-\frac1 4\times(x-5)^2\\ y-8=m\times(x-5)\end{align*}\end{rcases}}
have exactly one ordered real number pair as solution.
By expressing \latex{ y } from both equations, making the expressions obtained equal, performing the squaring, and then rearranging the equation so that one side is equal to \latex{ 0 } we obtain the following:
\latex{x^2 + (4m - 10)x + 37 - 20m = 0.}
This equation has exactly \latex{ 1 } solution when its discriminant is \latex{ 0 }, i.e.
\latex{(4m-10)^2-4\times(37-20m)=0.}
After rearranging the expression on the left-hand-side and after performing the possible simplifications, we get
\latex{m^2 - 3 = 0,}
which implies
\latex{m_1=\sqrt{3},m_2=-\sqrt{3}}.
The following straight lines passing through the point \latex{ P } have \latex{ 1 } point in common with the given parabola:
\latex{\begin{array}{lcl}t:\;x=5,\\ e_1:\;y=\sqrt3\times x+8-5\times\sqrt3,\\e_2:-\sqrt3\times x+8+5\times\sqrt3.\end{array}}
The straight lines \latex{e_1} and \latex{e_2} are each other's mirror images about \latex{ t }, and they have one point in common with the parabola so that all the other points of the parabola apart from the common point are in one of the half-planes defined by them. \latex{e_1} and \latex{e_2} are the tangent lines of the parabola passing through \latex{ P } (Figure 69).
\latex{ E }
\latex{ e }
\latex{ t }
Figure 70
The solution of example \latex{ 5 } showed that the concept of the tangent of a parabola is not as simple as the concept of the tangent of a circle, where it was enough to say that it has a single point in common with the circle.
Now we will give two precise definitions for the tangent of a parabola (Figure 70).
DEFINITON: The tangent of a parabola is a straight line, which has one point in common with the parabola, but it is not parallel with the axis of the parabola.
DEFINITION: The tangent of a parabola is a straight line, which has one point in common with the parabola, and all the other points of the parabola are in one of the half-planes defined by the straight line.
\latex{ F }
\latex{ e }
\latex{ d }
\latex{ t }
Figure 71
It can be proven that the two definitions are equivalent.
The tangent line parallel with the directrix and passing through the vertex of the parabola is the tangent at the vertex of the parabola (Figure 71).
Exercises
{{exercise_number}}. Set up the equation of the parabola with the vertex at the origin, which has a focal parameter of \latex{ 3 }, and the axis of which is parallel with the
- \latex{x}-axis;
- \latex{y}-axis;
Give the coordinates of the focus and also the equation of the directrix.
{{exercise_number}}. Give the focus, the vertex and the directrix of the parabola that has an equation of
- \latex{y=x^2};
- \latex{y=(x-2)^2};
- \latex{6y-12=-(x+1)^2};
- \latex{y=\frac{(x-1)^2}{4}}.
{{exercise_number}}. Calculate the coordinates of the common points of the parabola with the equation \latex{y^2 = 5 – x} and the straight line \latex{ e }, if the equation of \latex{ e } is
- \latex{x-y=1};
- \latex{2x+y+2=0};
- \latex{3x-4y=5};
- \latex{y=\frac{x}{2}-2}.
{{exercise_number}}. Give the equation of the tangent line passing through the point \latex{ P } of the parabola with the equation \latex{y = (x -2)^2 -2} the first coordinate of which is
- \latex{2};
- \latex{1};
- \latex{-3};
- \latex{0};
- \latex{5};
- \latex{-6}.
{{exercise_number}}. Give the equation of the straight line that touches the parabola with the equation \latex{y = (x + 3)^2 – 2} and passes through the point \latex{ P }, if
- \latex{P(0;0)};
- \latex{P(-3;6)};
- \latex{P(-2;-1)};
- \latex{P(-3;-2)};
- \latex{P(5;-2)};
- \latex{P(1;-6)}.
{{exercise_number}}. Calculate the coordinates of the common points of the parabola with the equation \latex{y = x^2} and the circle\latex{ k }, if the equation of \latex{ k } is
- \latex{x^2+y^2=4};
- \latex{x^2+(y-2)^2=25};
- \latex{x^2+(y+3)^2=9}.
{{exercise_number}}. One vertex of a regular triangle is the origin, its other two vertices lie on the parabola with the equation \latex{y = -x^2.} Calculate the coordinates of the missing vertices along with the perimeter and the area of the triangle.