https://www.urbanpro.com/ritam-r # Ritam R.

## IIT JEE Coach

## Overview

## Languages Spoken

## Education

## Address

## Verified Info

## Reviews

## FAQs

## Answers by Ritam R. (1)

## Answers by Ritam R. (1)

Ritam R.

Sakinaka East, Mumbai, India - 400072

Sakinaka East, Mumbai, India - 400072.

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

Details verified of Ritam R.✕

Identity

Education

Know how UrbanPro verifies Tutor details

Identity is verified based on matching the details uploaded by the Tutor with government databases.

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.

Hindi

English

Bengali

Indian Institute of Technology

Bachelor of Technology (B.Tech.)

Sakinaka East, Mumbai, India - 400072

Phone Verified

Email Verified

Report this Profile

✕

x

Is this listing inaccurate or duplicate? Any other problem?

Please tell us about the problem and we will fix it.

Engineering Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

this is test message this is test message this is test message this is test message this is test message this is test message this is test message

No Reviews yet! Be the first one to Review

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.

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

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

Like 0

Answers 16 Comments Engineering Entrance Coaching classes 2.5

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

MBA Entrance Coaching classes 2.5

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Class 11 Tuition 3.2

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 3.2

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

this is test message this is test message this is test message this is test message this is test message this is test message this is test message

No Reviews yet! Be the first one to Review

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

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

Like 0

Answers 16 Comments Share this Profile

Also have a look at

- Engineering Entrance Coaching classes
- MBA Entrance Coaching classes
- Class 11 Tuition
- Class 12 Tuition
- IIT JEE Coaching
- BITSAT Coaching
- Engineering Courses
- CAT Coaching
- MBA Courses
- Math Tutors
- Home Tutors
- Mathematics Tuition
- CBSE Tuition classes
- Private Tutors
- Tuition Tutor
- Home Tuition
- Private Tuitions
- Online Tutors
- Home Tuition for Class 12

X

Post your Learning Need

Let us shortlist and give the best tutors and institutes.

or

Send Enquiry to Ritam R.

Let Ritam R. know you are interested in their class

Reply to 's review

Enter your reply*

Your reply has been successfully submitted.