 Hello and welcome to the session. In this session we will discuss a question which says that if T is a variable parameter, rule says X is equal to A into square root of 1 plus T square. This whole over 1 minus T and Y is equal to A into a root 2T. This whole over 1 minus T lies on a fixed hyperbola. This is the value of the eccentricity of the hyperbola. Now before starting the solution of this question we should know our result. And that is the standard form of the equation of hyperbola is given by X square over A square minus Y square over B square is equal to 1. Now for the rectangular hyperbola length the transverse conjugate axis are equal that means 2A is equal to 2B so this means A is equal to B. Therefore the equation of the rectangular hyperbola by putting B is equal to A and this equation of hyperbola is this Y square is equal to A square. Now this result will work out as a T idea for solving out this question. And now we will start with the solution. Now in the question it is given that if T is a variable parameter we have to prove that the point XY and here the values of X and Y are given to us lies on a fixed hyperbola and also we have to find the value of the eccentricity of the hyperbola. So given X is equal to A into square root of 1 plus T square this whole upon 1 minus T and Y is equal to A into square root of 2T this whole upon 1 minus T is equal to square root of 1 plus T square whole upon 1 minus T Y over A is equal to square root of this hyperbola Further implies X square over A square is equal to 1 plus T square whole upon 1 minus T whole square over A square is equal to 2T over 1 minus T whole square. Now square minus Y square over A square is equal to now putting the values we have 1 plus T square whole upon 1 minus T whole square minus 2T over 1 minus T whole square which is further equal to whole square find the numerator this is the formula of 1 minus T whole square so it will be 1 minus T whole square over 1 minus T whole square which is further equal to 1 therefore over A square is equal to 1 implies X square minus Y square is equal to A square hyperbola is X square minus Y square is equal to A square. Now let us name it as equation number 1 and we can write that the equation 1 represent now for the given points use of X and Y were given to us the equation of the rectangular hyperbola of a given hyperbola that is this hyperbola given by E square is equal to further gives B square is equal to A square into E square minus 1 the whole and here the city is greater than 1 now we have B square is equal to square minus 1 the whole but we have got a rectangular hyperbola in which A is equal to A in this equation we get A square is equal to A square into E square minus 1 the whole which implies A square over A square is equal to E square minus 1 which further gives 1 is equal to E square minus 1 and this implies 2 1 plus 1 which is 2 and which further gives E is equal to root 2 minus root 2 of the given question and that is all for this session enjoy the session.