Now assuming that the curve is y = f(x), we get the equation of the tangent at any point (x,y) on curve as

                          Y - y = dy/dx ( X - x)

 So the point at which tangent meets X axis is ( (x(dy/dx) - y)/(dy/dx)) , 0)

 and the point at which it meets Y axis is ( 0 , y - (dy/dx.)x )

So the area of triangle formed by the tangent and the co-ordinate axis is

 =  1/2 * ((x(dy/dx) - y)/(dy/dx))*(y - (dy/dx.)x) = 2 (given)

 So solving this equation we get a quadratic in dy/dx :

              x² (dy/dx)² - (2xy - 4)dy/dx + y² = 0

   Solving we get dy/dx = (2xy - 4 + 4(xy - 2)² - 4(xy)²) / 2 x²

             &          dy/dx = (2xy - 4 - 4(xy - 2)² - 4(xy)²) / 2 x²

    Solve these equations by putting xy = t and then using the boundary condition i.e the curve passes through (1,1).

cheers ......... let us know if u stuck somewhere.

 

Well the eccentricity is defined as the ratio of distance of any point on the curve from a fixed point to the distance from a fixed line.

 

In accordance to this definition e = 1means parabola.

 

We always use this definition and you can also use it without any hesitation.

 

cheers

hiiiiii

let us suppose x is cube root of unity.

 

               So x = 1^1/3  or x³ = 1.

 

               Now x³ = 1

                 x³-1 = 0

                 (x-1)(x² + x + 1) = 0 

 

                   Solving we get x = 1 , (-1 + i3 )/2 , (-1 - i3 )/2

 

I hope it helps.

 

cheers

 

Hii

We take a general point say (h,k) and satisfy the condition given in the question and then replace (h,k) by (x,y) to get the equaation of curve which will give the locus.

I now this would have not made much sense to you .. but what I suggest is that u should post a question .... so that I know that the problem is in what sort of questions. I will solve the question for you and then probably the things would be more clear. Also soon we will have material on co-ordinate goemetry up on this very site which will guide u on solving problems.

 

cheers

The rank is defined as the number of linearly independent columns in a matrix.

Do you understand what linear independence means? If yes you can check for yourself that null matrix have no column which is linearly independent.

 

This way you can find rank of any matrix. I think this helps you.

 

Well the S L Loney book is suppose to be the best. It will guide you through all the basic concepts of the topic.

Feel free to ask more if u do not like the book!!!!!!

Hmm if u talk about the coaching material of Narayna then I can say that Narayana's material is compilation of all the other packages that other coaching institutes provide. So it is a good material. I have seen it and can say that it is worth reading. However here I must comment that our study material will be soon on the site and it is definitely much better than any material which any of the institutes provide.

 

Now about the faculty. The faculty is also good, infact many of the FIITJEE faculty has joined Narayna. But u must remember that it is u urself who would make the difference. Any teacher can just help, rest is all on ur hard work. :) 

The book for Algebra I suggest is S.K. Goyal from Arihant Publications. It is a very good book in my view. Very easy to follow and u can find it anywhere easily.

 

I hope this advice helps u !!! :)

So to solve this problem we suppose that the 3 consecutive nos are n, n+1, n+2. These numbers are a,ar^p and ar^q respectively (i.e. we have assumed that the 3 consecutive numbers can form terms of G.P.

 

So now we see that (n+1)/n = r^p and (n+2)/n = r^q .

 

                       ((n+1)/n)^q = ((n+2)/n)^p = r^pq

 

                       (n+1)^q = (n+2)^p. n^(q-p)                 ... (1)

 

 Now we claim that no two consecutive integers can divide each other except that one is 1 or -1.

 

Let us see why?

The two numbers be n and n+1.

So if n divdes n+1 then n+1/n is integer i.e 1/n is integer which is possible only if n is 1 or -1.

 

n cannot be -1 because then n+1 will become 0 and zero cannot be term of a GP.

 

Also if n = 1 then eq 1 becomes

 

                  2^q   = 3^p which is impossible because 2 and 3 are primes and they have no factor in common. So how can two numbers which have no common factors be equal.

 

Again returning to eq 1. Now we have this fact that n+1 cannot divide n+2 as well as n.

 

So now the LHS of eq1 when divided by n+1 will give an integer whereas the RHS will not and thus eq 1 cannot hold.

 

This clearly means that three consecutive numbers cannot form 3 terms of G.P.

 

I hope I have cleared your doubt ........ even if doubt persist write back.

 

 

 

Well the triangle would be divided in how many parts would depend on the fact that whether the n points chosen include the end points of third line or not ? If we assume that the end points are excluded then we can divide the triangle in n+1 parts. But how do we reach this soultion? well let us start with n = 1. Now if we join that point with the opposite vertex then we have divided the triangle in two (1 + 1) parts. Similarily if we further visualize we can see that n points will result in n+1 parts.