Make Ellipse Equation

Under the picture that relates the foci, eccentricity, and length of the main and negligible axes. Derive the identities if we remember that if T is a point on the ellipse the addition of the distances from T to the two foci is a steady.  Derive the equation for the ellipse with vertical main axis.  The derivative for the horizontal main axis case is similar.

Vertical Ellipse and Horizondal Ellipse

Making ellipse equation:

Consider the upside of the ellipse.  The addition of the distances from the two foci to the up side of the make a ellipse is

(b –c) + (b + c) = 2b

The square of is given by

4b2

Then consider the right hand side of the ellipse.

See that the distances from each side to the right hand side is the equal.  The square of sum is

(2d)2 = 4d2

Using the Pythagorean Theorem,

d2 = a2 + c2

Replacement and setting these same to each other gives

4b2 = 4(a2 + c2)

So that

b2 = a2 + c2

Place the equation in the particularly simple form

`(x^2)/(a^2) + (y^2)/(b^2) =1 (or) (x^2)/(b^2) + (y^2)/(a^2) =1`

Make ellipse equations - Examples:

Make ellipse equations - Example1:

Given the following equation

9x2 + 4y2 = 36

a) Solve the x and y intercepts of the graph of the equation.

b) Solve the co-ordinates of the foci.

c) Solve the length of the main and slight axes.

d) Make the graph of the equation.

Solution:

9x2 / 36 + 4y2 / 36 = 1

x2 / 4 + y2 / 9 = 1

x2 / 22 + y2 / 32 = 1

a = 3 and b = 2 (NOTE: a >b) .

Set y = 0 in the equation get and Solve the xaxes

x2 / 22  = 1

Solve for x.

x2  = 22

x = ± 2

Set x = 0 in the equation get and Solve the y axes

y2 / 32 = 1

Solve for y.

y2  = 32

y = ± 3

b) We need to solve c first.

c2 = a2 - b2

a and b were found in part a).

c2 = 32 - 22

c2 = 5

Solve for c.

c = +- (5)1/2

The foci are    F1 (0, (5)1/2) and  F2 (0 , -(5)1/2)



c) The main axis length is given by  2 a = 6.

The slight axis length is given by  2 b = 4.

d) The graph of the equation is


Example Ellipse

Equivalence Properties of Equality Tutor

Tutor is nothing but the teacher or instructor give notes to the students who comes through online. Tutor teaches a certain topic to student what they want by online. Here tutor give instruction about the equivalence property of equality to students by online. Equality represents that both side of the equation is of equal status or not. To represent this equality; we use the equivalence property of equality.

Equivalence Properties of Equality Tutor:

There are three properties under the equivalence of equality. These properties are applicable for the arithmetic operation.

Reflexive property:

Reflexive property is said to be any number that is equal to same number.

Example: a = a

Symmetric property:

Symmetric property is said to a number x is equal to the number y; then the number y is equal to the number x.

x = y then y = x

Transitive property:

Transitive Property is said to be a number an equal to the number b then the number b equal to the number c; then we can say the number an equal to the number c.

a = b; b = c and then a = c

Example Problems – Equivalence Properties of Equality Tutor:

Example 1:

Verify the following statement is transitive Property

M + 3 = 7; 7 = 4 + 3 and then 4 + 3 = 7

Solution:

Given: M + 3 = 7; 7 = 4 + 3 and then 4 + 3 = 7

Let take M + 3 = 7 ---- > equation 1

7 = 4 + 3 ---- > equation 2

4 + 3 = 7 ----- > equation 3

Step 1: Find the value of M in equation 1.

M + 3 = 7

Step 2: Subtract the number 3 on each side of the equation 1, we get

M + 3 – 3 = 7 – 3

M = 4

Thus 4 + 3 = 7 --- equation 4

That the equation 1 and equation 4 are equal,

4 + 3 = 4 + 3

Thus we proved the given statement is transitive property of equivalence.

Example 2:

Which of the following is the symmetric property?

a) m = n and n = m

b) m = 1/m and m= m

c) m + 1 = m - 1

d) m = 1 + m

Solution:

Symmetric property: x = y; y = x

Here the option ‘’a’’ refers the symmetric property.

m = n and n = m

Answer: a

These are the example problem istruct by the tutor about the equivalence properties of equality.

Fraction in Containers

Fractions are method of writing numbers that are not  whole numbers. Some fractions are written as  `1/2` ,`16/41` ,`3/2`. Fractions contains of two numbers, numerator and denominator. In the general form,

Numerator
Fraction = ----------------
Denominator

Where, numerator is at  top and denominator on the bottom.

For example,

In the fraction `2/3` , 2 is numerator and 3 is denominator. Now we see the problems for solving a fractions.


Conditions for fractions:

Some of the conditions are there for fraction addition and subtraction is given as,

To add the fractions the denominator must be same, then the numerators can is counted directly as same as numbers.

If the fractions the denominator is not same means, make them same number by multiplying and dividing by same scale factor.

Fraction subtraction is similar to the fraction addition. Let us see some of example problems using containers with same denominator and various denominators in detail and the example problems for containers fractions for better understand.

Example Problems for fraction in containers:

Example 1:

Find the total amount of water in containers; two containers are filled with one half of water and other containers have the capacity of eight liter in which four liter is filled.

Solution:

In first containers water filled is one half i.e `1/2.`

In second containers water filled is 4litre of total 8 liter capacity i.e `4/8`

The fractions are `1/2` and `4/8`

`4/8` can also be written as `1/2` (divide my 2 on both numerator and denominator.)

Total amount of water = `1/2 + 1/2`

= `(1+1)/2`

= `2/2`

= `1/1`

= 1

Example 2:

Find the total amount of water in containers; two containers are filled with one half of water and other containers have the capacity of eight liter in which six liter is filled.

Solution:

In first containers water filled is one half i.e `1/2.`

In second containers water filled is 4litre of total 8 liter capacity i.e `6/8`

The fractions are `1/2 ` and `6/8`

Total amount of water =` 1/2 + 6/8`

= `1/2 * 4/4 + 6/8`

= `4/8 + 6/8`

= `(4+ 6)/8`

= `10/8`

= `5/4`


Example 3:

Find the total amount of water in containers; two containers are filled with one half of water and other containers have the capacity of three liter in which one liter is filled.

Solution:

In first containers water filled is one half i.e `1/2.`

In second containers water filled is 4litre of total 8 liter capacity i.e `1/3`

The fractions are `1/2` and `1/3`

Total amount of water = `1/2 + 1/3`

= `1/2 * 3/3 + 1/3 * 2/2`

= `3/6 + 2/6`

= `5/6`

Sum of two Distributions

In sum of probability distributions and statistics, a probability will make the value of a random variable  or the probability of the value drawing  within a particular interval. The sum probability distribution defiines the measure of possible values like random variable can accomplish and the probability sum that the value of the random variable is within subset of that range.

Probability distributions:

Random Experiment and Sample Space

• Any experiment cannot be predicted in before, but is one of the set of possible outcomes, which is known as random experiment.
• An experiment performed frequently, every repetition is known as trial.
• The collection of all likely outcomes in random experiment is denoted the sample space.
• The sample space of a random experiment is denoted as a set S that includes all possible outcomes of the experiment
•  For simple experiments, the sample space may be exactly the collection  of possible outcomes.
• The sample space is a mathematically best set which contains the possible outcomes and perhaps other elements as well.
•  For example, if the testing is to draw a standard die and evidence the outcome, the sample space is S 1 2 3 4 5 6 , the set of possible outcomes.

Probability Distributions
      Let S acting as sample space with random experiment.
     Then X : x₁ x₂ x₃ …x_n
      P(X) is known as probability distribution of x.
Mean and variance of  Addition of two distributions:
Mean is E(X) = ∑ pi xi
variance is given below,
 

Examples:

1)Find the sum of mean of two distributions for the given below?

X	0	1	2
P(X)= x	0.25	.20	.25

X	0	1	2
P(X)= x	0.15	.20	.35

Solution:
Mean for table 1.
Mean E(X)1 = ∑ pi xi
            = 0*0.25 + 1*0.20+2*0.25
                       =0 + 0.20 +0.50  = 0.70
           Mean for table 2.
Mean E(X) 2 = ∑ pi xi
                    = 0*.15+1*0.20+2*0.35=0+.20+0.70
                 = 0.90
Sum of two distributions are E(X)1 + E(X)2= 0.70+0.90 = 1.6
2)Find out sum the variance of the following two distributions?

X	0	1	2
P(X)= x	0.25	.20	.25

X	0	1	2
P(X)= x	0.15	.20	.35

Congruent tTiangles Worksheet

In geometry, two figures are congruent if they have the same shape and size. More formally, 2 sets of points are called congruent if, and only if, one can be transformed into the other by an isometry, i.e., a combination of translations, rotations and reflections.

(Source-Wikipedia)

Congruent triangles worksheet -Example 1:

Let PQRS be a parallelogram and PR be one of its diagonals. What can you say about triangles PQRC andRSP?

Solution:
Step 1: In a four-sided figure, opposite sides are congruent. Hence sides

Step 2: QR and PS are congruent, and also sides PQ and RS are congruent.
Step 3: In a four-sided figure opposite angles are congruent.

Step 4: Hence angles PQR and RSP  are  four side figure.
Step 5: Two sides and an included angle of triangle PQR are congruent to two corresponding sides and an included angle in triangle RSP. According to the above postulate the two triangles PQR and RSP are congruent.

Congruent triangles worksheet-Example 2:

To solve the  triangle ABC congruent to triangle EFV by SAS?

Solution:
Step 1:Segment AB is congruent
Step 2: To segment of ST is
Step 3:  AB = EF = 4.
Step 4: Angle B is congruent to
Step 5: angle T because angle B = angle F = 100 degrees.
Step 6:Segment BC is congruent to segment FV because BC = FV = 5.
Step 7: ABC is congruent EFV is a SAS.



Congruent triangles worksheet-Example 3:
To solve the angle is QR is congruent to angle ST.

Solution:
Step 1: The triangle RPQ is congruent to triangle RTS.             
Step 2: Angle P is congruent to angle S because that information is given  in the diagram.
Step 3: Segment PR is congruent to segment RS because that information is given in the diagram.
Step 4: Angle QRP is congruent to angle SRT because vertical angles are congruent.
Step 5: Triangle SPQ is congruent to triangle RTS by Angle-Side-Angle.
Step 6: The triangles arecongruent, you know that all orresponding parts must be congruent.
Step 7:By CPCTC, segment BC is congruent to segment CE.

Unit Circle Sine Cosine

The Unit circle is used to understanding the sins and cos of angles to find 90 degree triangle. It is radius is exactly one. The center of circle is said to be origin and its perimeter comprises the set of all points that are exactly one unit from the center of the circle while placed in the plane.

It s just a circle with radius ‘one’.

The distance from the origin point(x,y) is `sqrt(X^2+Y^2)` by using Pythagorean Theorem.

Here radius is one So, The expression should becomes sqrt(x²+y²) =1

Take square on both sides then the eqn. becomes,

                      X²+y² =1

Positive angles are found using counterclockwise from the positive x axis

And negative angles are found using  anti clockwise from negative axis.

To find sine and cosine angle:

Angles are measures using counterclockwise from the positive x axis.The point where the ray at angle theta meets the unit circle is,
                                      (cos(`theta` ),Sin(`theta` )).
Here,cos(`theta` )----->Represents notation for the cosine of theta
Sine(`theta` ) is the notation of the sine of theta.
X= cos`(theta)`
Y=sin`(theta)`
The angle measured in radians is the length of the circular arc from the positive x
Axis counterclockwise until the desired angle is reached.
Since the radius of the unit circle is 1, the length of its perimeter is
2pr = 2p(1) = 2p.
Hence the angle for going exactly around the circle once is 2 radians. This
corresponds to 360 degrees.

Example problem -Unitcircle sines and cosine

Example 1: 

                 The angle q=0 radians is found by starting at the positive x-axis and going a distance 0. Hence the ray at angle q = 0 meets the unit circle at (1,0).

Hence
(cos(0), sin(0)) = (1,0),
cos(0) = 1,
sin (0)= 0.
Example 2. Without a calculator, find cos(p) and sin(p).
Solution:

The angle q= p radians is found by starting at the positive x-axis and going a distance p counterclockwise around the circle. But 2p is the total perimeter, so going a distance p arrives exactly half-way around the circle. Hence the ray at angle q = p meets the unit circle at (-1,0).

Hence
(cos(p), sin(p)) = (-1,0),
cos (p) = -1,
sin (p)= 0.

Polynomial Inequality

Inequality is a statement which means that the two things are not equal. Polynomials are the expressions which have the degree of at least 2 and more. An inequality that has polynomials is known as polynomial inequality. And polynomial inequality is not linear anymore. Solving polynomial inequality is little more complicated when comparing with linear inequalities. We will see how to solve polynomial inequality in the following sections.

Steps to solve polynomial inequality:

The following are the steps involved in solving polynomial inequality

Step 1: Make the inequality to obtain a zero on one side of the inequality

Step 2: The next step is to factor the polynomial which is on the left side of the inequality sign

Step 3: Determine the points at where the polynomial is zero.

Step 4: Graph the points we get in the previous step that is where the polynomial is zero

Step 5: Write down the answer

Example problems on solving polynomial inequality:

1)Solve the polynomial inequality x2 +15 < 8x

Solution:

Step 1: Make the inequality to obtain a zero on one side of the inequality

x2 +15 < 8x

x2-8x+15<0

Step 2: factor the polynomial x2-8x+15

x2-8x+15

x2-3x-5x+15

x(x-3) -5(x-3)

(x-5)(x-3)

Step 3: Determine the points at where the polynomial is zero.

Our polynomial is (x-5)(x-3)

When x= 5 ,

(x-5)(x-3) = (5-5)(5-3)

= 0(2)

=0

When x=3

(x-5)(x-3) = (3-5)(3-3)

= (-2)0

=0

The polynomial gets zero at x=5 and x= 3

Step 4: Graph the points we get in the previous step x=5, x=3 ( where the polynomial is zero )




Step 5: Write down the answer.

The solution of the given polynomial inequality is 3<x<5 .

2) Solve the polynomial inequality (x+2)(x+3) >0

Solution:

Step 1: Make the inequality to obtain a zero on one side of the inequality

Here in our example already there is a zero on the right side of the inequality .

Step 2: Factor the polynomial (x+2)(x+3)

No problem. It is in factored form

Step 3: Determine the points at where the polynomial (x+2)(x+3) is zero.

(x+2)(x+3)

When x= -2 ,

(x+2)(x+3) = (-2+2)(3+3)

=(0)(6)=0

When x= -3,

(x+2)(x+3) = (-2+3)(-3+3) = 0

Our polynomial get 0 at x=-2, x=-3

Step 4: Graph the points we get in the previous step x=-2, x=-3 ( where the polynomial is zero )

Algebra is widely used in day to day activities watch out for my forthcoming posts on Common Factor and solving systems of linear equations by elimination. I am sure they will be helpful.

Step 5: Write down the answer.

The solution for the given polynomial inequality is  -3<x<-2