Hi Akash,

I am an NITian and not IITian. I can answer it in a better way because I have realized and I know what it takes to be an IITian.

In Inorganic chemistry you have to mainly memorize the facts, except in some cases. At initial stage you can refer to NCERT and memorize it thoroughly and ensure that not a single line is left unstudied. Then you should refer to some advanced text books. Keep memorizing periodically and keep revising each chapter.

Coordinate geometry is very much important for JEE advanced point of view. And it is easy to score in this topic of mathematics. Coordinate geometry is less about concept and more about how fast you can see the actual trick to solve the problem. Someone can take 2 pages to solve a particular question but a smart aspirant will solve it instantly. This is what works in JEE Advanced. At first you should solve your coaching notes questions and then past year paper. Now comes the actual work to be done for excelling in JEE Advanced. Now start solving a different book, probably TMH or SK Goyal will do. Solve as much questions as you can. If you find these questions easy then switch yourself to SL Loney and start the same.

The next mistake most people make in this section is calculation mistakes. Avoid this at any cost and this can be achieved by practicing and only practicing.

If you follow this then you are surely going to crack JEE advanced.

All the best.

Hey!

First clear your basic concepts of plane geometry and solve some basic questions from ncert, then try to solve the advanced level questions without looking at the solution. Once you are done with the practice of same, then solve the previous years problems.

For more information you can go through this link

https://www.google.com/amp/s/school.careers360.com/articles/how-to-prepare-for-kvpy/amp

Good Luck

Greetings Aspirant,

The Answer to your question is as follow:

Let the equation of the tangent be

y=mx+c

Since it passes through (2,8) hence

8=2m+c

Or

c=8−2m

Hence the equation of the tangent becomes

y=mx−2m+8

Substituting in the equation of the hyperbola

5x(Square) −y(Square) =5

5x(Square) − (mx−2m+8)(Square) = 5

5x(Square)−5−(m(Square)x(Square)+4m(Square)+64−4m(Square)x−32m+16mx)=0

x(Square)(5−m(Square) )+x(4m(Square) −16m)−4m(Square) +32m−69=0

Since it is a tangent hence D is 0.

Or

B(Square) −4AC=0

(4m(Square) −16m)(Square) −4(5−m(Square) ) (−4m(Square) +32m−69)=0

m=3 and m=23/3

Hence

y=3x+2 and 3y=23x−22

For Your Better Understanding i have also uploaded an image of the same answer to your question. You may go through it.

Answer Image (https://drive.google.com/drive/folders/17Q4BTcoBKGmANQIhhdKF_5ZZkJcqGWAP?usp=sharing)

Hope My Answer Helped You.

Best of Luck For Future.

Well the answer is yes. There will be a lot of geometry questions that will be asked in jee mains and advanced. You may expect questions of different patterns from the dame topic. Like find the area between two spheres, find the points of intersection, find the planes of intersection etc.

Hi Unnati,

If sides of a triangle are 6,10 and x, then for x = 8, the area of triangle will be maximum.

We can understand this, by hit and trial method to obtain this answer. As, this is the most relevant answer.

Feel free to ask doubts in the Comment Section.

I hope this information helps you.

Good Luck!