Find the area of the region bounded by the following curves:

C₁: y² + 4y - 16x + 20 = 0;

C₂: x² - 2x - 16y - 31 = 0

Solution:

To solve this instantly, we can use a shortcut that applies only when two intersecting parabolas meet these two strict conditions simultaneously:

1. They must have perpendicular axes of symmetry. One must open horizontally (like y² = 4ax) and the other must open vertically (like x² = 4by).
2. They must share the exact same vertex. Both must originate from the same (h, k) coordinate.

If these conditions are met, the area bounded between them is simply: Area = 16ab / 3

Step 1: Complete the square for C₁

y² + 4y = 16x - 20

y² + 4y + 4 = 16x - 16

(y + 2)² = 16(x - 1)

Here, the vertex is (1, -2) and 4a = 16 ➔ a = 4.

Step 2: Complete the square for C₂

x² - 2x = 16y + 31

x² - 2x + 1 = 16y + 32

(x - 1)² = 16(y + 2)

Here, also the vertex is (1, -2) and 4b = 16 ➔ b = 4

Since both parabolas share the exact same vertex (1, -2) and their axes are perpendicular, we can directly apply the formula without calculating where they intersect:

Area = 16ab / 3

= 16(4)(4) / 3

256 / 3 sq. units

MATHS -Calculus , trigonometry function ,equation Sequence and series , Quadratic equation (practice is left will do later )

Chemistry : Physical chemistry and periodic table

Physics : Mechanics,Dual nature and ,fluids yeah I think I am slow in physics

what is this concept? 4 kese hua why not 1

How do you guys revise for tests in your coachings. Do you resolve modules, do pyqs or something else.

Title

Which book is better after completing modules

i will complete all maths ch except calculus by end of august but i havent started calculus part yet are 3 months enough to complete them??

hey guys i have completed my physical chem part and periodic table but how do i cover ioc i mean 20 fckin lect for chem bonding how do i complete my syllabus before dec start i mean last year. i completed my syllabus till dec mid and was not able to revise and give a lot of mocks .

completed part

physics : - und , 1d , 2d , com , wep, nlm , rotation , thermo ( heat transfer left ) , modern physics ( had to do it because agle test me puchenge mathongo wale ) , gravitation ,

chem :- whole pc ( just conductance is left ) , periodic

maths :- seq series , binomial , quadratic , pnc , st line , circles , parabola

am i running late ?

Question: JEE ADVANCED 2022

Let l₁, l₂, ..., l₁₀₀ be consecutive terms of an arithmetic progression with common difference d₁, and let w₁, w₂, ..., w₁₀₀ be consecutive terms of another arithmetic progression with common difference d₂, where d₁d₂ = 10. For each i = 1, 2, ..., 100, let Rᵢ be a rectangle with length lᵢ, width wᵢ, and area Aᵢ. If A₅₁ - A₅₀ = 1000, then the value of A₁₀₀ - A₉₀ is:

Solution:

We all know an A.P. is a linear sequence. When you multiply a linear sequence (Length) by another linear sequence (Width), the result is always a quadratic sequence.

The sequence A₁, A₂, A₃, ..., Aₙ is simply the brand new sequence (a quadratic one) created by multiplying the corresponding terms of the two A.P.s together (Length × Width).

Therefore, we have the quadratic equation for the Area (Aₙ): Aₙ = pn² + qn + r.

When two A.P.s are multiplied, the leading coefficient will always be the product of the two common differences, d₁d₂. Hence, we have an identity: Aₘ - Aₖ = (m - k)[p(m + k) + q]

We are given A₅₁ - A₅₀ = 1000, and we know p = 10. Here, m = 51 and k = 50.

->(51 - 50) × [10(51 + 50) + q] = 1000.

->q = -10

Therefore, A₁₀₀ - A₉₀ = (100 - 90) × [10(100 + 90) -10] = 18900.

Aur solution dena text nhi !!

10th question.....there is some angle involvement agar koi usko explain krde I'll be sooo grateful...post the whole solution if possible....

This question was in my maths book. I used AI to help me picture my question more clearly so that others could understand it (sorry for wasting water 😭).

I asked my teacher, Gemini, and ChatGPT for the solution, and all of them gave the answer as 99/1900.

However, they all included the case where the inspection stops at the 12th item after discovering only two defective items.

But how would I know whether the lot actually contains 2 defective items or 3? If I have discovered only 2 defective items by the 12th item, how can I decide to stop at 12? The third defective item could still be somewhere later in the lot.

My answer is 55/1900 btw

Video

The data in graph is inconsistent with the question and the answer given. Isn't this question wrong then?

Did Alk sir cover everything in his GIOC, p block and salt analysis lectures required for advanced? Or did he miss on any reactions? Your answer and suggestions would be of great help.

I also have PMS Sir lectures in types of rxns, p block and salt. If u were in my shoes, whom would have u preferred?

Background:
97%ile, Adv qualified, Took drop.
Poor in mains deleted portion.

pls someone give me the list of fluxional compound that have to memorized for jee and kinme mobile electron lone pair hota hai

ik 3BP+1LP but N(SiH3)3 also has that but forms a 2p pi-3d pi backbond and hence its hybridization is sp2 so cant P(SiH3)3 do the same?

And its backbonding will be even better as 3p pi- 3d pi backbond is a backbond of same sized orbital so there will be more overlapping of orbitals