**Parabola:**

From parabolic telescope mirrors
to receivers, the parabola plays a major role, and so it is useful to study the
properties of this particular figure. Probably the classic parabola of form:

y = x

y = x

^{2 }
is a good place to start. A
graph of this equation is shown below:

Note the vertex is on the

**origin**or (0,0), but as with the circle and ellipse this need not be so, and one can encounter parabolas that are displaced, e.g. away from the x-axis, as well as oriented in a different way. For example, consider: x

^{2}

^{ }= - y or y = - x

^{2}

^{ }

One can also obtain parabolas
with vertices oriented to the left or right of the y-axis. It is left to
readers to try to find equations that yield these.

The general form of the parabola is often given as:

The general form of the parabola is often given as:

^{2}= - 4py

Provided the parabola opens downward as shown above. If the reverse is true it is given as: x

^{2}= 4py . In each case, the

*focus*of the parabola is on the axis of symmetry (e.g. follow the y-axis through the vertex downwards - or upwards in the earlier case) and p units from the vertex.)

Let's consider the parabola shown above for x

^{2}= - y

^{ }then:

- 4p = 1 and p
= - 1/4

That is, - 1/4 unit

Now, the directrix of the parabola will be the line y = p or y = -1/4 for

x

then the slope of the tangent at any point is dy/dx = x/ 2p. which is 0 at the origin. But the second derivative: d

Of course, not all forms for the parabola are as straightforward as shown. Consider the equation:

2 x

We want to try to get this into the more general form for the parabola:

(x - h)

We begin by dividing both sides of the first equation by 2:

x

Which can be written as:

x

We can then complete the square, i.e. adding (- 3/4)

x

Or, write as:

(x - 3/4)

We can factor out 5/2 from the right hand side, i.e.:

(x - 3/4)

Now, compare this to the general form:

(x - h)

From this we see that: h = 3/4, k = -23/40 and 4p = 5/2 so: p = (4)/ (5/2) = 5/8

Therefore, the following hold true and can be confirmed by sketching the graph. :

a) The vertex is situated at V(3/4, - 23/40)

b) The axis of symmetry as always is x = h or x = 3/4 in this case.

c) The focus is p units below the vertex at F(3/4, - 6/5)

That is, - 1/4 unit

*from the vertex*to get to the focus F. What would we have for the previous parabola? Well, you should be able to work out: p = 1/4Now, the directrix of the parabola will be the line y = p or y = -1/4 for

*the downward oriented*parabola. It is a line parallel to the x-axis and by that amount of displacement. It should therefore be no surprise that the tangent to a parabola at its vertex is parallel to the directrix. Consider a parabola of form:x

^{2}= 4pythen the slope of the tangent at any point is dy/dx = x/ 2p. which is 0 at the origin. But the second derivative: d

^{2}y / dx^{2}= 1/2p which is positive.(So the curve is concave upward).Of course, not all forms for the parabola are as straightforward as shown. Consider the equation:

2 x

^{2}+ 5y - 3x + 4 = 0We want to try to get this into the more general form for the parabola:

(x - h)

^{2}= - 4p (y - k)We begin by dividing both sides of the first equation by 2:

x

^{2}+ 5y /2 - 3x/2 + 2 = 0Which can be written as:

x

^{2}- 3x/2 = - 5y /2 - 2We can then complete the square, i.e. adding (- 3/4)

^{2}to both sides:x

^{2}- 3x/2 + 9/16 = - 5y /2 - 2 + 9/16Or, write as:

(x - 3/4)

^{2}= -5y/ 2 - 23/16We can factor out 5/2 from the right hand side, i.e.:

(x - 3/4)

^{2}= -5/ 2 (y + 23/ 40)Now, compare this to the general form:

(x - h)

^{2}= - 4p (y - k)From this we see that: h = 3/4, k = -23/40 and 4p = 5/2 so: p = (4)/ (5/2) = 5/8

Therefore, the following hold true and can be confirmed by sketching the graph. :

a) The vertex is situated at V(3/4, - 23/40)

b) The axis of symmetry as always is x = h or x = 3/4 in this case.

c) The focus is p units below the vertex at F(3/4, - 6/5)

**Hyperbola**

**:**

In the
2016 film

*'Hidden Figures'*the most prolific human computer (Katherine Johnson) often had to compute when an elliptical orbit became hyperbolic. In the case of the ellipse we saw that the equation for this curve in Cartesian coordinates was:
x

^{ 2 }**/**a^{2 }+ y^{ 2 }**/**b^{2}= 1__Note__: In working these problems it also helps to realize that the ellipse can't have an e -value greater than 1 (which would make it

*a hyberbola*) and one also can't have imaginary values for the radical defining c.

Recall the
equation relating semi-major (a) and semi-minor (b) axis to identify location
of the foci was:

c
= (a

^{2}- b^{2})
But in
the case of the hyperbola this will become"

b
= Ö (c

^{2}- a^{2})
Given now
c is greater than a.

The
equation for the hyperbola in Cartesian coordinates is:

x

^{2 }**/**a^{2 }- y^{2 }**/**b^{2 }= 1
A sketch
of a representative hyperbola:

(x

^{2 }**/**9^{2 }- y^{2 }**/**4^{2 }= 1)
Is given below:

As seen
here the hyperbola, like the ellipse, is symmetric with respect to both axes
and the origin but it has no real y-intercepts. In fact, no portion of the
curve lies between the lines x = a and x = -a, or in this case x = 3, and x =
-3. The equation of the asymptotes is also easily obtained, i.e. y = -bx/
a and y = bx/a, or y = -2x/ 3 and y = 2x/ 3.

As we see
from this example, the two foci (F1 and F2) lie on the x-axis. Obtaining the
coordinates of the foci is straightforward and is just: F2 (+c, 0) and
F1(-c, 0) where:

c
= Ö (a

^{2}+ b^{2}). In this case, c = Ö (9 + 4).
= Ö 13. F1 is then at (+ Ö 13, 0) and F2 is at

(- Ö 13 , 0) where

**Ö**13 = 3.6.We can also interchange x and y in the basic hyperbola equation, changing it to a hyperbola with its foci on the y-axis instead of the x -axis. In other words, we now write:

y

^{2 }**/**9^{2 }- x^{2 }**/**4^{2 }= 1The graph for this hyperbola is shown below, with the equations identified for the straight (asymptote) lines:

Note the
center of a hyperbola is the point of intersection of its axes of symmetry. For
greater generality, if the center is at (h, k) we can also introduce a
translation to new coordinates such that:

x' = x -
h and y' = y - k

with
origin O' at the center. Then in terms of the new (translated) coordinates, the
equation for the hyperbola will be one of the following:

1)
x'

^{2 }**/**a^{2 }- y'^{2 }**/**b^{2 }= 1
2)
y'

^{2 }**/**a^{2 }- x'^{2 }**/**b^{2 }= 1*Example:*

Use the
translation of axes technique to analyze the equation:

x

^{ 2 }- 4 y^{ 2 }+ 2 x + 8 y - 7 = 0
And
thence find the equation for a hyperbola with center at x' = 0, y' = 0 and also
identify the values of (x, y). Obtain the coordinates of the foci F1 and F2 and
sketch the curve, label the foci and show the asymptotes.

*Solution*:

As we did
before (ellipse, parabola) complete the squares in the x, y terms separately
and reduce to standard form, so:

(x

^{2 }**+ 2 x) + 4 (y**^{2 }- 2 y) = 7
(x

^{2 }**+ 2 x + 1) + 4 (y**^{2 }- 2 y + 1) = 7 + 1 - 4
Then:

(x +
2)

^{2 }**/**4 + (y - 1 )^{2 }= 1
Applying
translation of axes:

x' = x +
1, and y' = y - 1

Thus
reducing the equation to:

x'

^{2 }**/**4 - y'^{2 }**/**1 = 1
Which
represents a hyperbola with center at x' = 0 and y' = 0, or x = -1 and y = 1.

Then: c
= Ö (a

^{2}+ b^{2}). = Ö 5
The
sketch of the hyperbola is shown below with foci and asymptotes identified:

It can
easily be verified that the straight lines (asymptotes) have equations: y = x/
2 and y = -x/2

__Problems for the Math Maven:__1)Given V(-3, 1) and F (0, 1) find the equation of the associated parabola and also for its directrix. Sketch the graph showing the focus, vertex and directrix.

2) Find the tangent to the parabola y = x

^{2 }using the fact that the equation of the line tangent to a curve y = f(x) is given by:

y - y1 = f'(x1) (x - x1)

where (x1, y1) = P1 is the point of the curve for which the tangent is constructed and y1 = (x1)

^{2 }. Show the graph of the given parabola and the tangent line with the

*point P1 (x1, y1) identified*.

^{ 2 }

**- 4 y**

^{ 2 }- 2 x + 8 y - 2 = 0

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