Eccentricity of 1

Conic sections:
Conic sections are formed when a right cone is intercepted by a plane.The shape so formed depends on the angle at which the conic is cut.
Consider the below figure of right cone intercepted by plane and we can see the different shapes formed.



Conic sections:
Conic sections are formed when a right cone is intercepted by a plane.The shape so formed depends on the angle at which the conic is cut.
Consider the below figure of right cone intercepted by plane and we can see the different shapes formed.



The first picture formed is known as parabola
Second one is ellipse
The third is hyperbola.

explanation to eccentricity of 1

 Eccentricity of 1  :-
Eccentricity is the measure of the deviation of a conic from a circular path .The parabola is the conic haing eccentricity as 1.

Definition of parabola :-
Parabola is locus of points whose distance from fixed line, called directrix, and a from a fixed point ,called focus,is equal. Vertex is a point where parabola changes its direction .The distance from focus to vertex and vertex to directrix is equal.

The main compenents of Parabola are :-
i)Vertex
ii) axis
iii) focus
iv) directrix
Consider the parabola given below

Description  of paraboala :-
i)Point F is called the focus of parabola
ii)Point "O" is called as the vertex of parabola.
iii)C ,D are the points on the parabola whoce distance fron focus and directrix are equal .
iv) Line AB is called directric fo parabola
v) Line ox is called the axis of parabola.

i) Y² = 4aX                                                                      

This parabola opens towards right side ie + ve x axis having  vertex at  (0,0). and focal distance "a" and focal point (a,0) . This curve transforms to (Y-k)² = 4a(X-h) when vertex shifted to a point  (h,k). with focus shifting to (a+h ,k)

ii) Y²= -4aX

This parabola opens towards  left side ie -ve  x axis having  vertex at  (0,0). and focal distance "a" and focal point (-a,0).This curve transforms to (Y-k)² = - 4a(X-h) when vertex shifted to a point  (h,k) and focus shifts to (h-a ,k)

iii) X²= 4aY
This parabola opens towards  top  side ie +ve  y axis having  vertex at  (0,0). and focal distance "a"and focal point (0,a).This curve transforms to (X-h)² = 4a(Y-k) when vertex shifted to a point  (h,k) and focus shifts to (h,k+a).

iv) X²=-4aY
This parabola opens towards down side ie -ve  y axis having  vertex at  (0,0). and focal distance "a" and focal point (0,-a).This curve transforms to (X-h)² = - 4a(Y-k) when vertex shifted to a point  (h,k-a).

examples to eccentricity of 1

Ex 1:
Given the equation of parabola Y² = 8X find the focus of parabola
solution:
Comparing with standard form of equation
Y² = 4aX
we get  4a =8
=> a=2
so focus of parabola is (a,0) = (2,0)

Ex 2:
find focus of parabola   y =  1/4 x²
solution :
y =  1/4 x² 
=> x²= 4y
=> a= 1 on comparing with standard form
so focus of parabola is (0,1) as axis of symmetry is +ve y axis

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Ex 3:
Given a parabolic equation Y²= -16X-32
Find i)  Vertex of parabola
ii) Focus of parabola

Solution:
Given Y²= -16X-32
=>  Y²= -16(X+2)
=>  Comparing with standard equation (Y-k)² = - 4a(X-h)
(h, k) =(-2,0)
a=4
i) vertex of parabola is (h,k) =(-2,0)
ii)focus of parabola is (-6,0) as parabola is opening towards left side.

Draw isosceles Triangle

The triangle is a closed geometric shape that contains three sides. There are several types of triangles in the geometry world. We can divide the triangles according to their angles and sides. We can define the isosceles triangle by using its sides.

      The triangle that has two equal or congruent sides in its measure is called as isosceles triangle. We can say that the equilateral triangle is also as an isosceles triangle,since the equilateral triangle has three equal sides . while we draw a isoscles triangle we have to see that two sides triangle must be equal.

draw Isosceles Triangle:

• when we draw a isoceles triangle, angles are congruent i.e., an angles of each sides are equal.
• when we draw a isoceles triangle diagonals of an isosceles triangle must be congruent.
• The measurement of the adjacent angles is giving the 180 degree. x = y = 180^o

Formulas for Isosceles triangle:

• Hight,  h = √(b² - `(1/4)` a²)
• Perimeter of Isosceles Triangle = A + B + C
• Area of isosceles triangle A = (b* h)/2

Example problems:

1) Find the area of isosceles triangle with the base and height are 5 cm and 7 cm.

solution:

We can find the area of given problem using the following formula:

Area A= (b*h)/2

Substitute the values of b and h into the above formula. Then we get,

         A= (5*7)/2

 Here multiplying to the values of 5 and 6 then dividing by 2.

  Then we get the final solution.   

 Answer A=17.5 cm²

2) Find the perimeter of Isosceles triangle that has side, Side 1 =10 cm, Side 2 = 10, Side 3 = 7 cm.

Solution:

   Given, Side 1 =10 cm, Side 2 = 10, Side 3 = 7 cm.

  Perimeter of Isosceles triangle   = Side 1+ Side 2+ Side 3

                                                         = 10 + 10 + 7

  Perimeter of Isosceles triangle  = 27 cm

3) Find the height of the isosceles triangle of the base a = 7 cm and equal sides b = 11 cm.

Solution:

The height of the isosceles triangle is given by the formula:

             h = √(b² - `(1/4)` a²)

substitute the a = 7cm and b = 11 cm. in the formula,

             h = √(11² - (1/4)*7² )

             h = √(121 - `(49/4)` )

             h = √(121 - 12.25 )

             h = √(108.75) = 10.428 ≈ 10.4 cm

Height of the isosceles triangle = 10.4 cm

Straight Triangle

In geometry, a triangle is the fundamental shapes. The polygon with 3 corners and 3 sides are the line segments. The corner otherwise called as the vertices. The sides otherwise called as the edges. A triangle with corners A, B and C is represented by ∆ABC.Here we will see about the triangle types and formulas.

Classification of Straight Triangles

The straight triangle has three types. The names are as follows,

• Equilateral Triangle
• Isosceles Triangle
• Scalene triangle

Equilateral triangle
     Here all three sides are the equal length. An equilateral triangle is also a normal polygon with every angles calculate 60 degree. The equilateral triangle is as follows,
               
Isosceles triangle
     These triangle two sides only the equal length. The figure is as follows,
                 
Scalene triangle
Here all sides are the different type. The three angels are also various in measures. This triangle figure is as follows,

           

Classified Based by the Internal Angels

• A triangle does not have an angle that calculates 90 degree is known as the oblique triangles.
• It has all interior angles calculating less than 90 is known as the acute triangle or acute-angled triangle.
• A straight triangle has one angle that calculates more than 90 degree is known as obtuse triangle or obtuse-angled triangle.

Formula
Triangle area is A=(1/2).b.h
Base is denoted by b and height is denoted by h

Formula for angles
     Sin(q)=opposite/Hypotenuse
     Cos(q)=Adjacent/Hypotenuse
     Tan(q)=Opposite/Adjacent

Properties

• Angle sum property
• Exterior angle property
• Triangle Inequality property
• Pythagoras theorem

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Examples

1)Find the area of straight triangle with the base 5 cm and the height 7cm.
Solution
     Given b=5 cm and h=7 cm.
     Formula A=(1/2).b.h
                    =(1/2).5 cm.7cm
                    =(1/2).35cm²
                    =17.5cm²

2)what is the obtuse triangle with the base 4 inches and the height 6 inches.
Solution
     Given b=4 in and height h=6 in
     Formula A=(1/2).b.h
                    =(1/2).(4 in).(6 in)
                    =(1/2).24 in²
                    =12in²

Solve Regular Polygons

If all sides of the polygon is same length in all of its sides and all of its angles are equal, then the polygon is known as regular polygon.  The regular polygon cannot be a concave polygon. Most regular polygons are convex or star. We will see about Solve regular polygons in this article.


Regular polygon examples:
Few examples of Regular polygons are,
                      Square                                 Equilateral triangle                  Pentagon                                    
               

                        Hexagon

                    

To solve regular polygons formulas:

     The following formulas are used to solve regular polygons,

• Interior angle of each side = `(180(n-2))/n ` degrees 

• Exterior angle of each side = `360/n ` degrees 

• Diagonal=` (n(n-3))/2` 

• Area = n x area of triangle 

      = ½ * (apothem * perimeter)  (Or)

• Area  = ½ *( n* s*r) ( or )

    A= `(s2 n)/ (4tan (pi/n))`
    And there is lot of formulas for area.
 Where, n = number of sides, s = side length, r = radius or apothem

Solve regular polygons Example problems:

Example 1:
The regular hexagon has the apothem 7 cm and side is 5 cm. Calculate the area.
Solution:
Now, let us solve regular polygons using the first formula (above mentioned)
By the formula,
Area = (½) * (apothem) *(perimeter)

Perimeter of hexagon = Length of the side * Number of side
                                    = 5 * 6
                                    = 30cm
   Area of the hexagon = (1/2) * 7 * 30
                                     = 105cm²
Example 2:
The octagon has the apothem of 9 cm and the side length is 6 cm. Find its area.
Solution:
Given, n=8, s=6,  r = 9
    Let us solve regular polygons using the second formula of area (above mentioned)
           By the formula
                                    Area = (½) * n * s * r
                                            = (1/2) * 8 * 6 *9
                                            = 216cm²
Example 3:
A regular pentagon has the side of 5 inches. Determine its area.
Solution:
Given,
N= 5(pentagon), s= 5 inches
We can use the third formula now,
                                      Area= `(s^2 n)/ (4tan (pi/n))`
                                       = ` (5^2 * 5) / ( 4tan(pi/5))`

                                      = `125/ 2.906`
                                     =  43.01inches²   


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Example 4:
 An equilateral triangle has the side of 5inches length. Find its area and perimeter.
Solution:
Length = a= 5 inches
By formula,
     Area of equilateral triangle =   `sqrt(3)/4` a²  
                                                     = `sqrt(3)/4` * 25
                                                     = 10.825 sq.inches
    Perimeter of the triangle= a+b+c
                                               = 5+5+5

                                               = 15 inches.

Surface Normals

Let us see about surface normals. A surface normal also called the simple normal, to a flat surface is a vector that is vertical to that surface.A line normal to a flat, the normal factor of the force, the normal vector, etc. The concept of routine generalizes to orthogonality.A normal is used for computer graphics.

Calculating a surface normals

For a convex polygon such as a triangle, a surface usual will be calculated as the vector cross product of two (non-parallel) edges of the polygon.
For a plane given by the equation ax + by + cz + d = 0.
For a plane is represent the following equation r = a + αb + βc.
where a is a vector to get onto the level surface and b and c are non-parallel vectors lying on the plane, the normal to the plane defined is given by b × c .
For a hyperplane in n+1 dimensions equation given from following equation r = a₀ + α₁a₁ + α₂a₂ + ... + α_na_n,
where a₀ is a vector to get onto the hyperplane and a_i for i = 1, ... , n are non-parallel vectors two-faced on the hyperplane, the normal to the hyperplane can be approximated by (AA^T + bb^T) ^{− 1}b where A = [a₁, a₂, ... , a_n] and b is a random vector in the space not in the linear span of a_i.

Cross product partial derivation


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In surface normal,a surface S is given completely   as the set of points (x,y,z) it satisfying F(x,y,z) = 0, so the normal point (x,y,z) on the plane.

For a surface S given clearly as a function f(x,y) of the independent variables x,y (e.g., f(x,y) = a₀₀ + a₀₁y + a₁₀x + a₁₁xy).

The normals can be identified in at least two equivalent ways.

The first getting its contained from F(x,y,z) = z − f(x,y) = 0, from the normal follows readily.

Angle Sum Theorem

In a triangle the sum of all the three interior angles will be equal to 180^o.  If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the interior opposite angles. Here we are going to see about the angle sum theorem.

Understanding Angle Bisector Theorem is always challenging for me but thanks to all math help websites to help me out.

Angle sum theorem:

Angle sum theorem of a triangle is equal to two right angles, i.e., 180 degrees
Given:
ABC is a triangle
To Prove
Angle A + Angle B + Angle ACB = 180^o
Produce BC to D. Through C draw CE || BA.

Proof of angle sum theorem of a Triangle:

Statement	Reason
1. Angle A = Angle ACE	Alternate angles angles BA is parallel to CE
2. Angle B = Angle ECD	Corresponding angles BA is parallel to CE
3. Angle A + angle B = Angle ACE + Angle ECD	statements (1) and (2)
4. Angle A + angle B  = Angle ACD	statement (3)
5. Angle A + Angle B + Angle ACB = Angle ACD + Angle   ACB	adding Angle ACB to both sides
6. But Angle ACD + Angle ACB = 180o	linear pair
7. Angle A + Angle B + Angle ACB = 180 °	statements (5) and (6)

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Corollary of Theorem on Angle theorem of a Triangle

If one side of a triangle is produced, the exterior angle so formed is equal to the sum of the interior opposite angles.
Given:
In Triangle ABC, BC is produced to D.
To Prove Corollary of Sum of Angles of Triangle:
Angle ACD = Angle A + Angle B



Proof:

Statement	Reason
1. Angle ACB + Angle ACD = 180o.	linear pair
2. Angle A + Angle B + Angle ACB = 180o	sum of the angles of a triangle = 180
3.  Angle ACB + Angle ACD = Angle A + Angle B + Angle ACB	statements (1) and (2)
4. Angle ACD = Angle A + Angle B Reason	statement (3); Angle ACB is common

Hard Fraction Problems

Hard fraction problems:

This articles discusses hard fraction problems solving. A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (`1/2` , `5/8` , `3/4` etc.) and which consist of a numerator and a denominator. (Source – Wikipedia)

Hard fraction problems solve for proper fraction, improper fraction and complex fraction problems.

Example for Hard fraction problems:

Example 1:
Subtract the fractions `4/5` – `3/4`

Solution:
The denominator (bottom number) is different so we have to take least common denominator (lcd).
LCD = 5 x 4 = 20
`(4 xx 4)/ (5 xx 4)` = `16/20` and `(3 xx 5)/ (4 xx 5)` = `15/20`
`16/20 ` – `15/20`
The denominators are equals
So subtracting the numerator directly = `(16-15)/20`
Simplify the above equation we get = `1/20`
Therefore the final answer is` 1/20`

Example 2:
Subtract the mixed fractions for given fractions,` 4 7/5` – `5 8/5`

Solution:
The given two mixed fractions are `4 7/5` – `5 7/5`
We need convert to mixed fraction to improper fraction `27/5` – `33/5`
The same denominators of the two fractions, so
                                         = `27/5` – `33/5`
Subtract the numerators the 27 and 33 = 27 - 33 = - 6.
The same denominator is 5.
                                         = `-6/5`
The subtract fraction solution is -`6/5` .

Example 3:
Multiply the mixed fractions for given two fraction,` 4 2/4` x `5 2/6`

Solution:
The given two mixed fractions are `4 2/4` x `5 2/6`
We need convert to mixed fraction to improper fraction `18/4` x `32/6`
Multiply the numerators the 18 and 32 = 18 x 32 = 576.
Multiply the denominators the 4 and 6 = 4 x 6 = 24
                                         =` 576/24`
The multiply fraction solution is 24

Example 4:
Convert `16/ (5/4)` to a simple fraction and reduce.

Solution:
The given complex fraction `16/ (5/4)`
Can be written as `16/1 -: 5/4`
First we have to take the reciprocal of the 2^nd number, and then multiply with the second one
Reciprocal of `5/4 ` is `4/5`
`16/1` x `4/5`
Multiply the numerator and denominator
`(16 xx 4) / (1 xx 5)` = `64/5`
Therefore complex fraction solution is `64/5`

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Practice for hard fraction problems:

Problem 1: Convert `5/ (5/4)` to a simple fraction and reduce.
Solution: 4.
Problem 2: Adding the mixed fractions for given two fraction,` 4 2/3` + `5 2/3`
Solution: `31/3`
Problem 3: Subtract the mixed fractions for given fractions,` 4 7/3` – `5 8/3`

Solution:`-4/3`