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Amitabh Anal Class 11 Tuition trainer in Delhi

Amitabh Anal

Delhi Cantt, Delhi, India - 110010.

Verified 7 Students

Referral Discount: Get ₹ 500 off when you make a payment to start classes. Get started by Booking a Demo.

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Education

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Overview

I can teach every topic very easily and make it stronger for the students.

Languages Spoken

Hindi Proficient

English Basic

Education

IGNOU 2009

Master of Business Administration (M.B.A.)

Address

Delhi Cantt, Delhi, India - 110010

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Teaches

Class 11 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

12

Board

CBSE, IGCSE

CBSE Subjects taught

Mathematics, Physics

IGCSE Subjects taught

Physics

Taught in School or College

No

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

15

Board

CBSE

CBSE Subjects taught

Physics

Taught in School or College

No

Teaching Experience in detail in Class 12 Tuition

I am giving tuitions to class 12th grade from last 15 years. I am providing concept of subjects to the students . The concept of Physics and Mathematics raises the confidence of the student. And the confidence in subject matter is the key to score high quality performance in result as well as in carrier building.

Reviews (2)

5 out of 5 2 reviews

Amitabh Anal https://s3-ap-southeast-1.amazonaws.com/tv-prod/member/photo/6693604-small.jpg Delhi Cantt
5.0052
Amitabh Anal
A

Class 12 Tuition

"He is really a teacher. He have a capability to give the complete idea of every topic either in Mathematics or Physics. He just go into the concept of relevant topic. So I highly recommend him for Physics and Mathematics for class 11th and 12th. "

Amitabh Anal
V

Class 11 Tuition

"He is one of those rare teachers who give their students a liberty to think out of the box by themselves. He has the ability to develop interest regarding the toughest subjects in science. Sir is quite good at explaining topics in a detailed way. One of those legends who can help you to master your concepts regarding physics and mathematics, he can adjust himself according to a student's pace to clear his/her doubts. Sir teach in a very friendly way, explaining tips and tricks to memorize even the toughest theroms or formulas. His educational influence is so great that it actually enabled me to write this comment describing my own experience. Thank you so much sir for being my teacher "

Have you attended any class with Amitabh?

FAQs

1. Which school boards of Class 12 do you teach for?

CBSE and IGCSE

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach Class 11 Tuition and Class 12 Tuition Classes.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for 12 years.

Answers by Amitabh (24)

Answered on 08/10/2019 Learn CBSE/Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3

It is given that x = acos3θ and y = asin3θ. Therefore, the slope of the tangent at is given by, Hence, the slope of the normal at ...more

It is given that x = acos3θ and y = asin3θ.

Therefore, the slope of the tangent at  is given by,

Hence, the slope of the normal at

Answers 3 Comments
Dislike Bookmark

Answered on 08/10/2019 Learn Tuition

Yes By putting the value of x and y on the left side of an equation , it is equal to the value of right side of an equation.
Answers 294 Comments
Dislike Bookmark

Answered on 07/10/2019 Learn CBSE/Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3

The equation of the given curve is ay2 = x3. On differentiating with respect to x, we have: The slope of a tangent to the curve at (x0, y0) is. The slope of the tangent to the given curve at (am2, am3) is ∴ Slope of normal at (am2, am3) = Hence, the equation of the normal at (am2, am3) is... ...more

The equation of the given curve is ay2 = x3.

On differentiating with respect to x, we have:

The slope of a tangent to the curve at (x0y0) is.

 The slope of the tangent to the given curve at (am2am3) is

∴ Slope of normal at (am2am3) = 

Hence, the equation of the normal at (am2am3) is given by,

y − am3 =

Answers 2 Comments
Dislike Bookmark

Teaches

Class 11 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

12

Board

CBSE, IGCSE

CBSE Subjects taught

Mathematics, Physics

IGCSE Subjects taught

Physics

Taught in School or College

No

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

15

Board

CBSE

CBSE Subjects taught

Physics

Taught in School or College

No

Teaching Experience in detail in Class 12 Tuition

I am giving tuitions to class 12th grade from last 15 years. I am providing concept of subjects to the students . The concept of Physics and Mathematics raises the confidence of the student. And the confidence in subject matter is the key to score high quality performance in result as well as in carrier building.

5 out of 5 2 reviews

Amitabh Anal
A

Class 12 Tuition

"He is really a teacher. He have a capability to give the complete idea of every topic either in Mathematics or Physics. He just go into the concept of relevant topic. So I highly recommend him for Physics and Mathematics for class 11th and 12th. "

Amitabh Anal
V

Class 11 Tuition

"He is one of those rare teachers who give their students a liberty to think out of the box by themselves. He has the ability to develop interest regarding the toughest subjects in science. Sir is quite good at explaining topics in a detailed way. One of those legends who can help you to master your concepts regarding physics and mathematics, he can adjust himself according to a student's pace to clear his/her doubts. Sir teach in a very friendly way, explaining tips and tricks to memorize even the toughest theroms or formulas. His educational influence is so great that it actually enabled me to write this comment describing my own experience. Thank you so much sir for being my teacher "

Have you attended any class with Amitabh?

Answers by Amitabh Anal (24)

Answered on 08/10/2019 Learn CBSE/Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3

It is given that x = acos3θ and y = asin3θ. Therefore, the slope of the tangent at is given by, Hence, the slope of the normal at ...more

It is given that x = acos3θ and y = asin3θ.

Therefore, the slope of the tangent at  is given by,

Hence, the slope of the normal at

Answers 3 Comments
Dislike Bookmark

Answered on 08/10/2019 Learn Tuition

Yes By putting the value of x and y on the left side of an equation , it is equal to the value of right side of an equation.
Answers 294 Comments
Dislike Bookmark

Answered on 07/10/2019 Learn CBSE/Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3

The equation of the given curve is ay2 = x3. On differentiating with respect to x, we have: The slope of a tangent to the curve at (x0, y0) is. The slope of the tangent to the given curve at (am2, am3) is ∴ Slope of normal at (am2, am3) = Hence, the equation of the normal at (am2, am3) is... ...more

The equation of the given curve is ay2 = x3.

On differentiating with respect to x, we have:

The slope of a tangent to the curve at (x0y0) is.

 The slope of the tangent to the given curve at (am2am3) is

∴ Slope of normal at (am2am3) = 

Hence, the equation of the normal at (am2am3) is given by,

y − am3 =

Answers 2 Comments
Dislike Bookmark

Contact

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Amitabh Anal conducts classes in Class 11 Tuition and Class 12 Tuition. Amitabh is located in Delhi Cantt, Delhi. Amitabh takes Online Classes- via online medium. He has 15 years of teaching experience . Amitabh has completed Master of Business Administration (M.B.A.) from IGNOU in 2009. HeĀ is well versed in English and Hindi. Amitabh has got 2 reviews till now with 100% positive feedback.

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