 Hello and welcome to the session. In this session we will discuss a question which says that if t is a variable parameter, root that x is equal to a by 2 into t plus 1 by t the whole and y is equal to a by 2 into t minus 1 by t the whole lies on a fixed hyperbola. What is the value of the eccentricity of the hyperbola? Now before starting the solution of this question, we should know a result. Now the standard equation of the hyperbola squared over a square minus y square over b square is equal to 1 where a and b are the constants and where 2a is the length of the transverse axis and 2b is the length of conjugate axis and in rectangular hyperbola length of transverse axis is equal to the length of conjugate axis that is 2a is equal to 2b which means a is equal to b. Therefore equation of the rectangular hyperbola that is by putting b is equal to a in the equation of hyperbola is given by x square minus y square is equal to a square. Now this result will work out as a key idea for solving this question and now we will start with the solution. Now in the question which is given that t is a variable parameter so we have to put the lies on a fixed hyperbola and we have the eccentricity of the hyperbola. Now given to a by 2 into t plus 1 by t the whole into t minus 1 by t the whole. Now this implies that t the whole is equal to minus 1 by t the whole. This further by 4 is equal to 1 by 4 into t minus 1 by t is equal to now putting the variable a square this will be 1 by 4 into minus 1 by 4 into t minus 1 by t whole square. Now taking 1 by 4 common it will be 1 by 4 into now t plus 1 will be equal to t square plus 1 by t square minus 2 are cancelled with each other. So this is equal to 1 by 4 into 4 is equal to 1 implies this y square is equal to a square. Now we know that square is equal to a square so we can write equation for the point x square of a rectangular hyperbola that is rectangular hyperbola time to city of the given hyperbola which is given by this equation then e square is over transverse axis whole square now conjugate axis is equal to 2b and transverse axis is equal to 2a so on solving this we get e square is equal to 1 plus b square over a square which further gives b square is equal to a square into e square minus 1 the whole. And here for case of hyperbola the simplicity e is greater than 1. Now it is 1 the whole that we have written a rectangular hyperbola which a is equal to b therefore putting b is equal to a in this equation we have 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 implies 1 is equal to e square minus 1 this further gives e square is equal to 1 plus 1 which is 2 and this implies e is equal to the equation of the given question and that is all for this I have enjoyed the session.