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Engineering Entrance Coaching classes > Engineering Entrance Coaching in Mumbai > Ritam R.

Ritam R. Engineering Entrance trainer in Mumbai/>

Ritam R.

IIT JEE Coach

Sakinaka East, Mumbai, India - 400072.

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Overview

Ritam R. describes himself as IIT JEE Coach. He conducts classes in Class 11 Tuition, Class 12 Tuition and Engineering Entrance Coaching. Ritam is located in Sakinaka East, Mumbai. Ritam takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. Ritam has completed Bachelor of Technology (B.Tech.) from Indian Institute of Technology. He is well versed in Hindi, English and Bengali.

Languages Spoken

Hindi

English

Bengali

Education

Indian Institute of Technology

Bachelor of Technology (B.Tech.)

Address

Sakinaka East, Mumbai, India - 400072

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Teaches

Engineering Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

MBA Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics

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

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

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FAQs

1. Which classes do you teach?

I teach Class 11 Tuition, Class 12 Tuition, Engineering Entrance Coaching and MBA Entrance Coaching Classes.

2. Do you provide a demo class?

Yes, I provide a free demo class.

3. How many years of experience do you have?

I have been teaching for less than a year.

Contact

Answers by Ritam R. (1)

Answered on 13/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +1 CBSE/Class 12/Mathematics

2 and 3

Hello Archit, Please see the solutions below: 2. Let (h,k) be the midpoint of the chord. Then the equation of the chord with midpoint (h,k) is: (hx/3)-(ky/7) = (h^2/3) - (k^2/7) (Equation given by T1 = S1). 7x + y = 20 is the equation of the same line. Comparing the coefficients of the line... ...more
Hello Archit, Please see the solutions below: 2. Let (h,k) be the midpoint of the chord. Then the equation of the chord with midpoint (h,k) is: (hx/3)-(ky/7) = (h^2/3) - (k^2/7) (Equation given by T1 = S1). 7x + y = 20 is the equation of the same line. Comparing the coefficients of the line we get: h/(3*7) = -k/(7) = ((h^2/3) - (k^2/7))/20 So k/h = -1/3 Now the line whose equation we require to find passes through the origin and (h,k) So the equation of the line is y = (k/h).x Since k/h = -1/3, the equation of the line is y = (-1/3)x ---------------------------------------------------------------------------------------------------------------------------------------------- 3. The equation of the hyperbola can be written as (x^2/16) - (y^2/25) = 1 Let the point on the Hyperbola be (4secA, 5 tan A) The equations of the assymptotes are given by (x^2/16) - (y^2/25) = 0 Factorizing out the 2 stratight lines, we get the 2 equations as y = 5x/4 and y = -5x/4 Let Q be the point on y = 5x/4. Since the abscissa remains same, the x coordinate of Q is 4secA. So the y coordinate of Q is y = 5.4secA/4 = 5secA So Q: (4secA, 5secA). Now R lies on y = -5x/4. Solving similarly like we did for Q, the coordinates of R would be (4secA, -5secA). We also have P: (4secA,5tanA) So, PQ = 5(secA-tanA) and PR = 5(secA+tanA) Therefore, PQ.PR = 25.((secA)^2-(tanA)^2) For any angle A, ((secA)^2-(tanA)^2) = 1 Therefore, PQ.PR = 25. Let me know if you have any questions. Thanks
Answers 16 Comments
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Teaches

Engineering Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

MBA Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics

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

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

No Reviews yet!

Answers by Ritam R. (1)

Answered on 13/10/2016 Learn Tuition/Class XI-XII Tuition (PUC) +1 CBSE/Class 12/Mathematics

2 and 3

Hello Archit, Please see the solutions below: 2. Let (h,k) be the midpoint of the chord. Then the equation of the chord with midpoint (h,k) is: (hx/3)-(ky/7) = (h^2/3) - (k^2/7) (Equation given by T1 = S1). 7x + y = 20 is the equation of the same line. Comparing the coefficients of the line... ...more
Hello Archit, Please see the solutions below: 2. Let (h,k) be the midpoint of the chord. Then the equation of the chord with midpoint (h,k) is: (hx/3)-(ky/7) = (h^2/3) - (k^2/7) (Equation given by T1 = S1). 7x + y = 20 is the equation of the same line. Comparing the coefficients of the line we get: h/(3*7) = -k/(7) = ((h^2/3) - (k^2/7))/20 So k/h = -1/3 Now the line whose equation we require to find passes through the origin and (h,k) So the equation of the line is y = (k/h).x Since k/h = -1/3, the equation of the line is y = (-1/3)x ---------------------------------------------------------------------------------------------------------------------------------------------- 3. The equation of the hyperbola can be written as (x^2/16) - (y^2/25) = 1 Let the point on the Hyperbola be (4secA, 5 tan A) The equations of the assymptotes are given by (x^2/16) - (y^2/25) = 0 Factorizing out the 2 stratight lines, we get the 2 equations as y = 5x/4 and y = -5x/4 Let Q be the point on y = 5x/4. Since the abscissa remains same, the x coordinate of Q is 4secA. So the y coordinate of Q is y = 5.4secA/4 = 5secA So Q: (4secA, 5secA). Now R lies on y = -5x/4. Solving similarly like we did for Q, the coordinates of R would be (4secA, -5secA). We also have P: (4secA,5tanA) So, PQ = 5(secA-tanA) and PR = 5(secA+tanA) Therefore, PQ.PR = 25.((secA)^2-(tanA)^2) For any angle A, ((secA)^2-(tanA)^2) = 1 Therefore, PQ.PR = 25. Let me know if you have any questions. Thanks
Answers 16 Comments
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Contact

Ritam R. describes himself as IIT JEE Coach. He conducts classes in Class 11 Tuition, Class 12 Tuition and Engineering Entrance Coaching. Ritam is located in Sakinaka East, Mumbai. Ritam takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. Ritam has completed Bachelor of Technology (B.Tech.) from Indian Institute of Technology. He is well versed in Hindi, English and Bengali.

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