An Important Result For IIT-JEE Advance 2022 || Ellipse

[@dhairyaarora4243]

Sir please continue this advance series for 2023 as well....these questions are really helpful!

[@linxuser897]

Elegance in maths and physics problem solving is probably the most satisfying thing. Gives me goosebumps

[@anonymous-bi6ul]

It was awesome !!

[@Fighter_Believer_Achiever]

Thank you very much sir 👏

[@archit3327]

SIR SHORTCUT IF a=b then ellipse is circle then min dis will be zero as perpendiclar from centre met tangent at POC IN CIRCLE SO IN GENERAL CASE MIN DIS. SHOULD BE (a-b)^n where n must be 1 to satisfy dimensional formula of distance....

[@higherthangods]

Good one.

[@jineshpagaria3421]

What if it is k(a-b)^2? Then though dimensionally correct ,it would still be wrong .

[@theexplorer9012]

@@jineshpagaria3421 same thought

[@bhupeshdewangan2772]

I think you just went reverse and generalised this.....how can you just write b-a? I can't see an analogy here...explain me if I am unable to get it

[@insan_IITB]

No bro u r wrong.. It could be k(a-b) here k could be any positive integer ( dimension less) Analogy could also be... a²-b²/a It is also Dimensionally correct 😃 So u r wrong 😁

[@harshithgowda6227]

We must all take a second to appreciate the Efforts put in to bring these high quality videos, not just the explaination (which is exceptionally good) but also the diagrams, your voice too is so convincing sir 👍, just a humble request if time permits please upload as many questions as possible before JEE Advanced 2022 with concepts like these.

[@_aryandadhich]

great video🔥🔥

[@younus559]

Please continue this series bro This is helpful

[@cutegirlakanshabag2449]

Thank-you sir 😊

[@gokulr8755]

Nice method. I also solved it without finding coordinate of foot of perpendicular by just using Pythagoras. Say r is the dist of point on ellipse from origin, d is distance of foot of perpendicular from origin. We need to maximize x which is √(r^2-d^2). Just to clean up some mess square on both sides and take derivative. 2xx'=2rr'-2dd' x,r and d are functions of theta. Now we get rr'=dd'. r=√(a^2cos^2θ+b^2sin^2θ), d can b found out with formula for perpendicular distance of line from a point. In the end the result from the derivative becomes ab=b^2cos^2θ+a^2sin^2θ and thed distance^2 becomes (a^2-b^2)(cos2θ), cos2θ can be evaluated from the previous condition and we get our ans as |a-b|

[@HershO.]

Very nice method! Slight correction, we do not need mod in the answer as it is given that a > b.

[@shourya3099]

Damn this quality content

[@Abhi-kr6df]

7:19 and that'll be all

[@potu6534]

Sir thank you very much 😊😊😊😊😊😊

[@letsgetitdone2161]

most satisfying maths problem

[@aadityaraj996]

i think you know ALOK SIR he is also teach like u and do derivations in class and say to remember these result...

[@sparshsinghchundawat7761]

thanks again sir for such a brilliant method....HATS OFF

[@raosahebbade715]

I am 2022 jee aspirant, if I solve this problem, (1) find the tangent equation (2) then dist OR by perpendicular distant of tangent from origin. (3) OP by distance between two points. (4) (OP)^2 - ( OR)^2 = PR^2 (5) then differentiating it , find find thita. (6) by putting thita find PR maxi value. So long ..... Aapke dimagme yese ideas 🤯kaise ate hain🤔 please tell us ...

[@KM-om1hm]

Dimmag ka hi khel ha, usse pucho 😂

[@adityasaha2627]

AWSOME CONTENT THIS CHANNEL IS GOING TO ROCK VERY SOON😍🔥

[@arnabdaw2587]

Plz bring more such content sir, 🙏🏻

[@154bharshbhanushali9]

Hello Sir ! JEE 2022 Aspirant , Got 99.676 percentile in Maths in 1st attempt , I am done with the PYQ question , So thinking to solve little bit Sameer Bansal Calculus Book , Can u suggest me , what to solve actually ? Becoz solving whole book is not practically possible now

[@pogpieunited3379]

That book is too good i too am confused btw i got 98.5 percentile

[@l1mbo69]

How will someone else tell you that, it's not like Irodov where some questions are really easy and some really tough. Use your own judgement to see if a problem is new for YOU and/or if it looks tough (you'll have to do same in actual JEE advanced)

[@jjbb3714]

What was your total percentile by the wau

[@yowaimo890]

Given the time boundation your better off with grb you can either choose between black book or sameer bansal for calculus , pink book krne ka time mile so atleast refer through the soln. To get at least get the idea on how to approach such problems. Baaki there's not much time jo bhi krna h jldi crow.

[@mulamallaharshithreddy7285]

we can also use pythagorus for tri OPR right sir

[@nishantrathore5031]

Thanks a lot your questions will boost my jee advanced preparation.

[@vedanthaldia]

My sir also discussed the same question with the same method. Also many stuff he taught is similar to your methods. Are you guys related in some way or the other ? 😂

[@Maddy_2206]

Sir if we apply am gm on all the 4 quantities a2 b2 a2tan2theta and b2cot2theta that will be greater than equal to 4ab in that case answer will be different ??

[@RZMATHS]

In that case you will take fourth root ?? You have taken sqaure root check

[@Maddy_2206]

I have taken fourth root only (a2 ×b2×a2tansquaretheta×b2cot square theta )^1/4 =(a4b4)^1/4 =ab So sum greater than equal to 4ab Max distance will be (a2-b2)/root(4ab)

[@l1mbo69]

No because a^2+b^2>2ab so total is >4ab. But problem is equality case of a^2+b^2>2ab is only reached for a=b which we don't know is true, we need to find general result here for some fixed a,b. If we wanted min of d over all possible (a,b) then we would indeed get the right answer, which is zero (as a=b, a^2-b^2=0 in numerator). And we can crosscheck that with our general result as well, which was |a-b|: over all (a,b) its minimum is also zero, and also achieved for a=b

[@Hustlerr_io]

This ques idea from 1999 jee adv pro of that triangle mini or max area to found out 🤓

[@cbaadith380]

This is a cemgage illustration 😂

[@DeepakKumar-kw8ex]

❤️❤️❤️

[@tamimmolla7054]

who are you and where have you studied?????????????????????????

[@tamimmolla7054]

@Sachin Singh YEH I KNOW I WAS JUST CURIOUS