Parabola Math Problems

A parabola  is the set of all the points whose distance from a fixed point in the plane are equal  to their distances  from a fixed line in the plane.  The fixed point is called the focus and the fixed line is the directrix.

.directrix
Formula  for parabola ;-
(1) y2 = 4px  has symmetry around  x axis Directrix x + p = 0  Latus rectum  LL' is 4p vertex = (0,0)
(2) y2 = -4px has symmetry around the negative side of x-axis Directrix  x - p = 0 . LL' = 4p. Vertex =(0,0)
(3) x2 = 4 py has symmetry around the  positive y - axis Directrix y + p = 0.  LL' = 4p. Vertex =(0,0)
(4) x2 = -4py has symmetry around the negative y - axis. Directrix y - p = 0.  LL' = 4p. Vertex= (0,0)

Problems on Parabola:
Here is a parabola math problem for us  to do.
# Find the focus , vertex, axes, directrix  of the parabola y2 - 4x - 4y = 0
Sol:-
parabola
Step 1 : Let us bring the above given parabola equation to the parabola formula equation.
y2- 4x - 4y = 0
Step 2 : Add 4x to both sides + 4x             + 4x
We get                 y2 - 4y       =    4x
Step 3 :Complete the square on LHS we write y2 - 4y +4  = 4x + 4 ( we add 4 to both sides of the equation to balance it)
Step 4 : Now we get                                      (y  - 2 )2    = 4(x+1)
Step 5 : Let us compare the above equation to the general equation Y2 = 4 pX
we get Y = y-2  and X = x+1. Hence Y2 = (y -2)2  and 4px = 4(x+1) . Hence we get p = 1
Step 6 : Now let us set a table for both the values. Now x = X - 1 and  y = Y + 2
____________________________________________________________________________
Referred to X and Y axes                             Referred to x and y axes
____________________________________________________________________________
focus S = (p,0)                                          x = X - 1 = 1 -1 = 0; y = Y +2= 0 + 2 = 2
So focus is (0,2)
____________________________________________________________________________
vertex is (0,0)                                            x = X - 1 => 0 - 1 = -1 and y = Y + 2 = 0+2 =2
So vertex = (-1,2)
____________________________________________________________________________
axes Axis  of parabola Y =0                       Axis of the parabola is y = Y+2 = 0+2 =2
so axis  is y - 2 = 0
____________________________________________________________________________
equation of the directrix is                          equation of the directrix is  x + 1 = -1
x = -p that means X = - 1                           which gives us  x + 2 = 0
____________________________________________________________________________
Hence for the given parabola  y2 -4x - 4y = 0 we have (1)  Focus  as (0,2)  ;
(2)  Vertex is (-1,2)
(3)  Axis of the parabola is y = 2
and     (4) directrix is x + 2 = 0

Condition for a Line to be a Tangent for a Parabola:-
Let us find the condition for a line  lx + my + n = 0 to be a tangent to the parabola y2 = 4ax
Solution:-
The condition that a  line y = mx + c  to be a tangent to the parabola y2 = 4ax is that C = a/m...(1)
Step 1 : Let us write lx + my + n = 0 in the form y = mx + c
That gives us my = -lx - n
y =  [ -l]x + [ -n]
m        m
Step 2 : Now we have  y = mx + c   => y = (-l/m) x  + ( -n/m)  That is slope = -l/m and  y intercept C = -n/m
Hence  C =  -n/m  and m(slope) = (-l/m)
Step 3   But  C =  a / m (formula)
So we write  -n  =    a =   - am
m    (-l/m)       l
Step 4   Hence we get - n =  -am
m      l
Step 5 : Cross multiplying we get  -nl = -am2
Step 6 : That gives us nl = am2
Hence the condition that lx + my + n= 0 to be a tangent to the parabola y2  = 4ax is  nl = am2

Geometric Constructions Tutorial

Geometric constructions is a basic construction of geometric figures. In this article we can see the construction of a perpendicular bisector, construction of an angle and construction of a triangle. For the construction of geometric figures we use only the help of a ruler (straight-edge) and compass. Compass is generally used to draw arcs of particular distances. The distance is calculated with the help of a ruler.
Geometric Constructions

i) To bisect a given line segment

Given: A line segment AB

To construct: The perpendicular bisector AB

Construction:      a) With A as centre and radius equal to more than half of AB, draw two arcs, one on each side of AB.

b) With B as centre and with the same radius taken above draw two more arcs cutting the previous arcs                              at C and D

c) Join C and D cutting AB at O. CD is the perpendicular bisector of AB.

perpendicular bisector of a line segment

ii) To construct an angle of 60°

Construction: a) Draw a line segment AB of any suitable length.

b) With A as centre and any suitable radius draw an arc cutting AB at C.

c) With C as centre and the same radius AC, draw another arc cutting the former arc at D.

d) Join A and D and produce it to E.
construction of angle 60degrees

iii) To construct an angle of 30°

Construction: a)First construct an angle of 60° as shown above

b) With C and D as centre, draw two arcs of equal radii cutting each other at F

c) Join A to F.
construction of angle 30degrees

Construction of more Geometrical Shapes

iv) To construct a triangle having given the lengths of three sides.

Let the three sides be 3.6cm, 4cm and 5.1cm

Construction: a) Draw a line segment AB = 3.6cm

b) With A as centre and radius = 4cm, draw an arc.

c) With B as centre and radius = 5.1cm draw one more arc intersecting the former arc at C.

d) Join AC and BC. Then ΔABC is the required triangle.(Note: The construction fails when the sum of any two sides is less than the third side.)

Algebra is widely used in day to day activities watch out for my forthcoming posts on Combinations and Permutations Examples and Continuity of a Function. I am sure they will be helpful.

v) To construct a triangle, having given the lengths of any two sides and the included angle.

Let the length of the two sides be 3.9cm and 4.3cm and included angle be 60°

Construction:a) Draw a line segment AN = 3.9cm

b) Construct
c) From AX, cut off AC = 4.3cm

d) Join BC. Then ΔABC is the required triangle.

construction of a triangle, where two sides and included angle are given

Difficult Logic Problems

Problem: - Which term will replace the question mark in the series:

ABD, DGK, HMS, MTB, SBL,?

Solution: Clearly, the first letters of the first, second, third, fourth, and fifth terms are moved three, four, five, six and seven steps forward respectively to obtain the first letter of the successive terms. The second letters of the first, second, third, fourth and fifth terms are moved five, six, seven, eight and nine steps forward respectively to obtain the second letter of the successive terms. The third letters of the first, second, third, fourth and fifth terms are moved seven, eight, nine, ten and eleven steps forward respectively to obtain the third letter of the successive terms.

Thus the missing term is ZKW             (Answer)
Difficult Logic Problems next Set

Problem: - A child is looking for his father. He went 90 meters in the east before turning to his right. He went 20 meters before turning to his right again to look for his father at his uncle’s place 30 meter from this point. His father was not there. From there, he went 100 meters to his north before meeting his father in a street. How far did the son meet his father from the starting point?

Solution: - Clearly the child moves from A 90m eastwards up to B, then turns right and moves 20m up to C, then turns right and moves 30m up to D. Finally, he turns right and moves 100m up to E.

difficult logic problem

Clearly, AB = 90m, BF = CD = 30m.

So, AF = AB- BF = 60m

Also, DE = 100m, DF = BC = 20m

So, EF = DE- DF = 80m.

Therefore, his distance from starting point A = AE = `sqrt[ (AF)^2 +(EF)^2]`  = `sqrt[(60)^2 + (80)^2]`

= `sqrt(3600 + 6400)` = `sqrt10000` = 100m    (Answer)

Problem: - Each odd digit in the number 5263187 is substituted by the next higher digit and each even digit is substituted by the previous lower digit and the digits so obtained are rearranged in the ascending order, which of the following will be the third digit from the left end after the rearrangement?

Solution: - After performing operation on the digit we get 6154278

Arranging the above number in ascending order we get 1245678

Here third digit from the left end is 4.                (Answer)
More Difficult Logic Problems

Problem: - In a certain code TEMPORAL is written as OLDSMBSP. How is CONSIDER written in that code?            (Answer: RMNBSFEJ)

Problem: - In a certain code language ‘how many goals scored’ is written as ‘5 3 9 7’; ‘many more matches’ is written as ‘9 8 2’ and ‘he scored five’ is written as ‘1 6 3’. How is ‘goals’ written in that code language?  (Answer: either 5 or 7)

Problem: - Reaching the place of meeting on Tuesday 15 minutes before 08.30 hours, Jack found himself half an hour earlier than the man who was 40 minutes late. What was the scheduled time of the meeting? (Answer: 8.05 hrs)

Number of Diagonals in a Pentagon

Polygon is any shape, which is enclosed by its sides. The line segments that connect any two vertices are called as diagonals. The names of the polygon are classified on the basis of their number of sides. Any polygons which possess five sides are called as pentagon. In this article, we shall discuss about the number of diagonals of pentagon. Also we shall solve problems regarding number of diagonals of pentagon.

Formula - Number of Diagonals of Pentagon:

The number of diagonals of any polygon can be obtained by using the formula,

 [n(n-3)]/2

where n is the number of sides of the polygon.

diagonals of pentagon

Now we shall determine the number of diagonals of pentagon.

The number of sides of a pentagon is five.

Therefore n = 5.

By substituting n = 5 in the above formula, we can determine the number of diagonals of pentagon.

Number of diagonals of pentagon = [5(5-3)]/2

= [5(2)]/2

= 10/2

= 5

Therefore the number of diagonals of pentagon is 5.


Example Problem - Number of Diagonals in a Pentagon:

Determine the number of sides and name of the polygon whose number of diagonals is 5.

Solution:

Given:

The number of Diagonals = 5

The formula to determine the number of diagonals is

[n(n-3)]/2 ,

where n is the number of sides of the polygon.

We have to find n:

Equate the both.

[n(n-3)]/2 = 5

Multiply by 2 on both sides:

2[(n(n-3))/2] = 2(5)

n(n - 3) = 10

Multiply by n within the bracket:

n2 - 3n = 10

Subtract 10 on both sides.

n2 - 3n - 10 = 0

n2 - 5n + 2n - 10 = 0

n(n - 5) + 2 (n - 5) = 0

(n - 5) (n + 2) = 0

n - 5 = 0 and n + 2 = 0

n = 5 and n = -2

The sides of polygon must not be negative. So ignore -2.

Therefore the number of sides of the polygon is 5.

Pentagon is the polygon which possess 5 sides.

Hence pentagon is the polygon which possess 5 sides and 5 diagonals in it.

Plane Distance Calculator

In a plane distance calculator we have the axes points such as (x,y) and this coordinates are placed on the quadrant, there are four quadrants for (x,y) axis. Origin is the center or starting point for the (x,y). In two dimension coordinate system we have the two x and y plane.We can measure the distance between plane by the formulas.In this article we have the formulas and the problems for finding the plane distance.

Plane Distance Calculator:

Distance between plane calculator can be measured by the following formualas and that can explained by the figur shown below. From the above figure we can clearly understand that the distance 'l' from the origin we can find the plane distance by the coordinates given.

D =  ` |d1-d2|/sqrt(A^2 + B^2 + C^2)`

distance between two plane

How to use plane distance calculator:

First seperate the d1,d2 and all othher parameters

Now, enter the parameters in the respective column

And get the result in the new column

Problems in Plane Distance Calculator:


Example 1:
Find the distance from the two plane 2x – 3y + 3z = 12 and –8x + 12y – 12z = 24.
Solution:

First seperate the d1,d2 and all othher parameters
Now, enter the parameters in the respective column
And get the result in the new column

2x – 3y + 3z = 12
–8x + 12y – 12z = 24.  by dividing equation by 4   we get 2x - 3y + 3z = -6.
First sepearte the
Formula for find the distance between the plane

D =  ` |d1-d2|/sqrt(a^2 + b^2 + c^2)`

Here, a =2, b= -3 and c=3  d1= 12 d2 =-6

= | 12 - (-6) | / √(4 + 9 + 9)

= 18/√22

Example 2:

Find the distance between the parallel planes z = x + 2y + 1 and 3x + 6y - 3z = 4.

Solution:

First seperate the a,b,c,d and all othher parameters

Now, enter the parameters in the respective column

And get the result in the new column

Formula for find the distance between the plane

D =  ` |ax_1+by_1+cz_1+d|/sqrt(a^2 + b^2 + c^2)`

Here, a =3, b= 6 and c=-3 d =-4

=  ` |3 xx 0+6xx0+(-3)xx1-4|/sqrt(3^2 + 6^2 + -3^2)`

= `7/sqrt54 `

Calculating Areas of Figures

In day-to-day life,we often came across the word area.In math, Area is nothing but a region bounded by a closed curve. In differential geometry of surfaces, area is considered as an important invariant.

Calculating areas of different figures is an important and an interesting one. In this article of  calculating areas of figures, the areas of different figures are calculated using the formulas.

Triangle:            Area of Triangle    =    ½ b h

b ---> base

h ---> vertical height

Rectangle:        Area of  Rectangle  =  l w

l ----> length

w ----> width

Square:               Area of a square   =  a2

a ----> side length

Parallelogram: Area of Parallelogram = b × h


b ----> breadth

h ----> height

Circle:                       Area of a Circle = pi r2

r ----> radius
Worked Examples for Calculating Areas of Figures:

Example 1:

Find the area of a triangle with base of  13 m and a height of 6 m.

Solution:

Area of a triangle =  ½ b h

=  ½ (13) (6)

=  39 m2

Example 2:

Find the area of rectangle given the length is 10 cm and width is 5 cm.

Solution:

Area of a Rectangle  =  l * w

= 10 * 5

= 50 cm2

Example 3:

Find the area of a square of side length 21 cm

Solution:

Area of a square  = a2

= 212

= 441 cm2

Example 4:

Find area of a parallelogram through base of 23 cm and a height of 17 cm.

Solution:

Area of a Parallelogram = b h

= (23) · (17)

= 391 cm2

Example 5:

The radius of a circle is 27 inches. Find its area.

Solution:

Area of  Circle = pi r2

= 3.14 (27)2

= 3.14 (729)

= 2289.06 in2
Practice Problems for Calculating Areas of Figures:

1) Calculate the area of rectangle given the length is 10 m and width is 7 m.

Answer: 70 m2

2) Find area of a square of side length 22 cm.

Answer: 484 cm2

3) Find the area of triangle given base is 14 cm and height is 7 cm.

Answer: 49 cm2

4) Find area of a parallelogram through base of 35 cm and a height of 15 cm.

Answer: 525 cm2

Radical Math Problems and Solutions

Radical problems and solutions are defined as one of the important topic in mathematics. Basically, there are three values are present in the radical number. Those values are named as called the index number, radical number, and the another one is known as the radicand number. For example, root(4)(12) is denoted as radical numbers. In this example of radical number, 4 is called as the index number, 12 is called as the radicand number. Mainly square root and the cubic roots are present in the radical statement.

Radical Expressions Calculator

The explanation for radical math problems and solutions are given below the following,

We can do many of the operation by using the radical. They are called as,

Addition problems and solutions by using the radical.
Subtraction problems and solutions by using the radical.
Multiplication  problems and solutions by using the radical.
Division problems and solutions by using the radical.

Example Problems and Solutions for Radical Math

Addition problems and solutions by using the radical.

Example 1: Add the following radical numbers, 12( sqrt(5) + sqrt(2) ) + 10( sqrt(2) + sqrt(5) )

Solution:

The given number is 12( sqrt(5) + sqrt(2) ) + 10( sqrt(2) + sqrt(5) )

= 12sqrt(5) + 12sqrt(2) + 10sqrt(2) + 10sqrt(5)

= 12sqrt(5)+ 10sqrt(5)  + 12sqrt(2)  + 10sqrt(2)

= 22sqrt(5) + 22sqrt(2)

This is the answer for radical numbers addition.

Subtraction problems and solutions by using the radical.

Example 2: Subtract the following radical numbers, 12( sqrt(5) + sqrt(2) ) - 10( sqrt(2) + sqrt(5) )

Solution:

The given number is 12( sqrt(5) + sqrt(2) ) - 10( sqrt(2) + sqrt(5) )

= 12sqrt(5) + 12sqrt(2) - 10sqrt(2) - 10sqrt(5)

= 12sqrt(5) - 10sqrt(5)  + 12sqrt(2)  - 10sqrt(2)

= 2sqrt(5)  + 2sqrt(2)

This is the answer for radical numbers subtraction.

Problem 3: Multiply the following radical numbers, 12( sqrt(5) + sqrt(2) ) and  10( sqrt(2) + sqrt(5) )

Solution:

12( sqrt(5) + sqrt(2) )   xx   10( sqrt(2) + sqrt(5) )

= 12sqrt(5) + 12sqrt(2) xx   10sqrt(2) + 10sqrt(5)

= 12 sqrt(10) xx  10 sqrt(10)

= 120 sqrt(100)

= 120  xx 10

= 1200

This is the answer for radical numbers multiplication.

Example 4: Divide the following radical numbers 1/(sqrt(7) - sqrt(8)) .

Solution:

1/(sqrt(7) - sqrt(8))

1/(sqrt(7) - sqrt(8))  xx  (sqrt(7) + sqrt(8))/(sqrt(7) + sqrt(8))

(sqrt(7) + sqrt(8))/(sqrt(7)^2 - sqrt(8)^2)

(sqrt(7) + sqrt(8))/(7 - 8)

(sqrt(7) + sqrt(8))/ - 1

= - (  sqrt(7) + sqrt(8) )

= - sqrt(7)  - sqrt(8)

This is the answer for radical numbers Division.
Practice Problems and Solutions for Radical Math

Example 1: Add the following radical numbers, 24( sqrt(3) + sqrt(12) ) + 12( sqrt(12) + sqrt(3) )

Answer: 36 sqrt(3)  +  36  sqrt(3)

Example 2: Subtract the following radical numbers, 24( sqrt(3) + sqrt(12) ) - 12( sqrt(12) + sqrt(3) )

Answer: 12 sqrt(3)  +   12 sqrt(3)