Parabola Equations in Algebra
These are my notes on parabola equations and graphs.
Conic sections are named after the different ways that a plane can intersect a
cone. The three basic conic sections are parabolas, ellipses, and hyperbolas. A
circle is an example of an ellipse.
A parabola with vertex (0,0) can be represented symbolically by the equation
\(y=ax^2\). With this representation, a parabola can open either upward when a>0
or downward when a<0.
A parabola is the set of points in a plane that are equidistant from a fixed
point and a fixed line. The fixed point is called the focus and the fixed line
is called the directrix of the parabola.
Example 1
Find the equation of the parabola with focus (-1.5,0) and directrix x=1.5.
Solution:
A parabola always opens toward the focus and away from the directrix. the
parabola should open to the left. It follows that p<0 in the equation
\(y^2=Apx\). The distance between the focus at (-1.5,0) and the vertex (0,0) is
1.5 and so \(p=-1.5<0\). Note that the vertex of the parabola is (0,0) because
the vertex always lies midway between the focus and directrix. The equation of
the parabola is \(y^2=4(-1.5)\)x.
Reflective Property of Parabolas
When a parabola is rotated about its axis, it sweeps out a shape called a
paraboloid. Paraboloids have a special reflective property. When incoming,
parallel rays of light from the sun or distant stars strike the surface of a
paraboloid, each ray is reflected toward the focus. If the rays are sunlight,
intense heat is produced, which can be used to generate solar heat. Radio
signals from distant space also concentrate at the focus. Scientists can analyze
these signals by placing a receiver at the focus. This property of a paraboloid
can also be used in reverse. If a light source is placed at the focus, then the
light is reflected straight ahead.
Example 2
Locating the receiver for a radio telescope
The US naval research laboratory designed a giant radio telescope weighing 3450
tons. Its parabolic dish has a diameter of 300 feet and a depth of 44 feet. find
an equation in the form \(y=ax^2\) that describes a cross section of this dish.
If the receiver is located at the focus, how far should it be from the vertex?
Solution:
Locate a parabola that passes through (-159,44) and (150,44). substitute either
point into \(y=ax^2\).
\[y=ax^2 \rightarrow 44=a(150)^2 \rightarrow a=44/150^2 = 11/5625\]
The equation of the parabola is \(y=\frac{11}{5625}x^2\).
The value of p represents the distance from the vertex to the focus. To
determine p write the equation in the form \(x^2=4py\). Then,
\(x^2=\frac{5625}{11}y\).
It follows that \(4p=\frac{5625}{11} \rightarrow p=127.84\). Therefore, the
receiver should be located 127.84 feet from the vertex.
Translations of Parabolas
If the equation of a parabola is given by either \(x^2=4py\) or \(y^2=4px\),
then its vertex is (0,0). We can use translations of graphs tof ind the equation
of a parabola with vertex (h,k) rather than (0,0). This translation can be
obtained by replacing x with (x-h) and y with (y-k).
\((x-h)^2 = 4p(y-k)\)
\((y-k)^2 = 4p(x-h)\)
Example 3
Graph the parabola with vertex (h,k)
Graph the parabola given by the equation \(x=-1/8(y+3)^2+2\).
Solution:
Rewrite the equation in the form \((y-k)^2 = 4p(x-h)\)
\(x=-1/8(y+3)^2+2\)
\(x-2 = -1/8(y+3)^2\)
\(-8(x-2) = (y+3)^2\)
\((y+3)^2 = -8(x-2)\)
it follows that the vertex is (2,-3) 4p=-8 or p=-2 and the parabola opens to the
left. The focus is located 2 units to the left of the vertex and the directrix
is located 2 units to the right of the vertex. Therefore, the focus is (0,-3)
and the directrix is x=4.
Example 4
Finding the equation of a parabola with vertex (k,k)
Find the equation of the parabola with focus (3,-4) and directrix y=2.
Solution:
The focus and the directrix are shown where the parabola opens downward (p<0),
and its equation has the form \((x-h)^2=4p(y-k)\). The vertex is located midway
between the focus and the directrix, so its coordinates are (3,-1). The distance
between the focus (3,-4) and the vertex (3,-1) is 3, so p=-3<0. The equation of
the parabola is \((x-3)^2 = -12(y+1)\).
Example 5
Finding the equation of a parabola
Write \(2x=y^2+4y+12\) in the form \((y-k)^2=(x-h)\)
Solution:
We can write the given equation in the required form by completing the square:
\[2x=y^2+4y+12\]
\[2x-12 = y^2+4y\]
To complete the square on the right side of the equation, add \((4/2)^2=4\) to
each side.
\[2x-12+4 = y^2+4y+4\]
\[2x-8 = (y+2)^2\]
The given equation is equivalent to \((y+2)^2 = 2(x-4)\)