 Hello and welcome to the session. In this session we will discuss a question that says that if even and e2 are the eccentricities of a hyperbola and its conjugate respectively, find e1 is to e2. Now before starting the solution of this question, we should know a result. And that is if x square over a square minus y square over b square is equal to 1, with a given hyperbola, then minus x square over a square plus y square over b square is equal to 1. That is y square over b square minus x square over a square is equal to 1 or you can write x square over a square minus y square over b square is equal to minus 1 and b are the constants. That is the hyperbola will be the conjugate and transverse axis of the given hyperbola is called the conjugate hyperbola of the given hyperbola. This is the eccentricity hyperbola then e square is equal to 1 plus which is equal to 1 plus now the length of the conjugate axis is 2b and length of transverse axis is 2a or square which is equal to 1 plus b square over a square. Now this result will work out as a key idea for solving out the solution. In the question it is given that e1 and e2 are the eccentricities of the hyperbola and its conjugate respectively then we have to find e1 is to e2. Now let minus y square over b square is equal to 1 with a given hyperbola then using the result for finding out the conjugate hyperbola of the given hyperbola. Here we have our b square minus equal to 1 e1 is the eccentricity that is the hyperbola then using this result which is given the key idea 1 square will be equal to 1 plus b square over a square. e1 is equal to square root of b square over 2 plus b square over the given hyperbola that is the conjugate hyperbola of the given hyperbola therefore e2 square is equal to 1 a square over b square this will be equal to 2a and the length of transverse axis will be equal to 2d. In this solving we are getting e2 square is equal to 1 plus a square over 2 square root of b square plus a square whole upon. Now we have to this to e2. Now using the values of e1 this to e2 is equal to square root of a square plus b square whole upon a square b square plus a square whole upon b square which is further equal to square root of a square into square root of b square over square root of b square plus a square. Now these terms are cancelled with each other. This is equal to square root of b square over a square. Therefore, e1 is to e2 is equal to square root of b square over a square, e1 is to e2 is equal to plus minus b over a, that's a given question and that's all for this session. Hope you all have enjoyed the session.