Identifying Conics

We identify conics with the following rules:

1. If only one of x and  y is squared, then the equation is a parabola.

Parabolas:

• If the square belongs to the x then it is a cup (positive coefficient) or an upside down cup (negative coefficient)

• If the square belongs to the y then it is a sideways parabola (positive coefficient opens right and negative coefficient opens to the left)

2.  If the signs of the x2 and y2 coefficient are different then it is a hyperbola

Hyperbolas:

• If the negative sign belongs to the y2 then the hyperbola crosses the x-axis

• If the negative sign belongs to the x2 then the hyperbola crosses the y-axis

3. If the coefficients are the same it is a circle

4. Otherwise it is an ellipse

Remark:  Always make sure that all of the variables are on the same side of the equation before you attempt to classify.

Examples

x2 -2y2 +x - y  =  14

We see that this is a hyperbola since the coefficients have opposite signs.

Exercises

Fill in the following table with either parabola, circle, or ellipse.  You can check your answer by putting your mouse over the box for a few seconds.

 Equation Type x             y2                         +                =   1                               1             9 x2             y2                         -                =   1                              4             4 x2                     y2                         =  1 +                                      16                   25 x2             y2                         +                =   1                              2              2 x2              y                         =                 + 4                              16            36

Compilation of Notes for Midterm I

Completing the Square Step by step process of completing the square:

Example

2x2 - 8x + 2

1. Factor the leading coefficient:  2(x2 - 4x + 1)
2. Calculate -b/2:  -(-4)/2 = 2
3. Square the solution above:  22  = 4
4. Add and subtract answer from part three inside parentheses: 2(x2 - 4x + 4 - 4 + 1)
5. Regroup:  2[(x2 - 4x + 4 ) + 3]
6. Factor the inner parentheses using part two as a hint:  2[(x - 2)2 +3]
7. Multiply out the outer constant:    2(x - 2)2 + 6
8. Breath a sigh of relief.

Some pointers on solving word problems.

2)  Label variables:  Write "Let x = ..."  where x should be a numerical variable that will give the solution.  Then label any other variables if needed

3)  If possible draw the picture.  Recall the Pythagorean Theorem:  a2 + b2 = c2  where a and b are the lengths of the legs of the right triangle and c is the length of the hypotenuse.

4)  Reread the problem phrase by phrase, inserting the variables when they are spoken in the phrase.  (Recall that is ,will be, was, are, equate to =.

5)  Using the phrases come up with appropriate equations.

6) Solve the equations.

1)  Isolate the radical so that the radical is alone on the left side of the equation with everything else on the other side of the equation.

2)  Square both sides of the equation

3)  Bring everything back to the left side of the equation so that the right side is a 0.

4)  Solve the resulting equation by factoring, the root method, the quadratic equation, of some other method.

5)  Plug your answer back into the original equation making sure that your result is an actual solution and not an extraneous one.

Often we encounter an equation that is a quadratic in disguise.  For example the equation

x4  + 3x2  -  4 = 0

is a fourth degree equation.  Notice that the exponent of the leading term is twice the exponent of the second term.  Because of this we can use what is called a substitution:

Let u = x2 then x4  = u2

Then the equation in terms of u becomes:

u2   + 3u - 4 = 0

which is a quadratic.  Hence, we can factor:

(u - 1)(u + 4) = 0

So that u = 1 or u = -4

Since u = x2  we have that

x2  = 1 or x2  = -4

The first has solution x = 1 or x = -1  and the second has no solution.

We will solve using the following:

1)  Put everything on the lift hand side so that we have for example Quad > 0

2)  Factor and set equal to zero.

3)  Solve and place answers on a number line.  This will cut the number line into two or three regions.

4)  Pick a test value for each region and plug that test value into each of the factors.  Put plusses or minuses over the region depending on whether the test values test positive or negative.

5)  If the region has two plusses or two minuses the the region is positive

If the region has one of each then the region is negative.

6) If the inequality is "<" then include the negative regions

If the inequality is ">" then include the positive regions

7) If the inequality is a less (greater) than or equal to then include the endpoints with solid dot and the interval []

If the inequality is a less (greater) than then do not include the endpoints by showing an open dot and the interval ()

To solve rational inequalities, we can use the same technique that we used for  quadratic inequalities with the following adjustments:

A)  After putting everything on the left hand side we put the left hand side over a comon denominator.

B)  Instead of factoring and solving to find the cut points, we just set the numerator and denominator each equal to zero.

C)  The cut point that is determined from that denominator will never be included, thus will be bordered by a ( and shown as an open dot.

Recall that a circle with radius r and center (h,k) is defined by the set of point of distance r from the point (h,k).  If (x,y) is on the circle then the distance from (h,k) to (x,y) is r.  The distance formula tells us that

(x - h)2  + (y - k)2 = r2

Example:  sketch the graph of the parabola y =  x2 - 4x -5

Step 1:  Find the x-coordinate of the vertex-  -b/2a

x = 4/2 = 2

Step 2:  Find the y-coordinate of the vertex by plugging the x coordinate into the equation.

y = 22 -4(2) - 5 = -9

Therefore the vertex has coordinates (2,-9)

Step 3:  Find the y-intercept by plugging in 0 for x

y = 02 - 4(0) - 5 = -5

Step 4:  Find the x-intercepts by setting y = 0 and factoring or quadratic formula.

0 = x2 - 4x - 5

(x - 5)(x + 1) = 0

x = 5 or x = -1

Step 5:  If necessary, plug in more values of x to find a few additional points.

(Here it is not needed since we already have four points:  (2,-9), (0,-5), (5,0), (-1,0)

If steps 1 through 4 produce fewer than 3 points, it is recommended to plot a few additional points.

Step 6:  Graph it!

Hence we have the standard form of an ellipse centered at the origin:

x2 /a2+  y2 /b2 = 1

The points (a,0), (-a,0), (0,b), and (0,-b) are called the vertices of the ellipse.

To plot the hyperbola with equation  x2/a2 - y2/b2 = 1 we follow these steps:

1)  Plot the intercepts (a,0), (-a,0) and the points (0,b), (0,-b) as with the ellipse.

2)  Draw a rectangle containing these four points.

3)  Draw the lines that contain the diagonals of the rectangle (these are the asymptotes)

4)  Draw the hyperbola that with the vertices (a,0) and (-a,0) that has the drawn asymptotes.