 Hello and welcome to the session. In this session we will discuss a question which says that in the hyperbola x minus 2 whole square over 81 minus y plus 1 whole square over 16 is equal to 1, find a part coordinates of center and eccentricity, b part coordinates of okai and vortices equation of x's and length of x's and d part length of latest vector. Now before starting the solution of this question we shall know some results. And that up for the hyperbola a square minus y square over b square is equal to 1 where a and b are the constants the coordinates of center 0 0 then the coordinates of vertices a 0 then the coordinates 0 where e is the eccentricity e is equal to square root of plus b square whole square a square the equation of the transverse x y is equal to 0 and the length of the transverse x is equal to 2a and the equation of the conjugate x's is x is equal to 0 and the length of the conjugate x's is equal to 2p and now let us start with the next part and that is the length of the latest vector 2b square over 8. Now these results will work out as a key idea for solving out this question and now we will start with the solution. Now the equation of the hyperbola is given to us so given the equation of the hyperbola is 2 whole square over 81 minus y plus 1 whole square over 16 is equal to 1. Now shifting the origin to 2 minus 1 by putting capital X is equal to x minus 2 and capital Y is equal to y plus 1 in this equation let it be equation number 1 we get capital X square by 81 minus capital Y square by 16 is equal to 1. Now this is the equation of the hyperbola with respect to the new axis. Now comparing it with the standard equation of hyperbola here a square is equal to 81 and b square is equal to 16 so this is the equation of the form x square over a square minus y square over b square is equal to 1 which is a horizontal hyperbola. Now in the first part we have to find the coordinates of center and eccentricity. Now using these results which are given in the key idea now the coordinates of center given hyperbola with respect to the new axis are capital X is equal to 0, capital Y is equal to 0 that is 0 0 and the coordinates of center to the whole x minus 2 is equal to 0 and y plus 1 is equal to 0 that is 2 minus 1. Now the eccentricity b is equal to square root of a square plus b square whole upon a square which is equal to square root of now a square is 81 b square is 16 so this is square root of 81 plus 16 whole upon 81 which is equal to square root of 97 by 81 which is equal to root 97 by 9 line. Now we know that eccentricity of a hyperbola is greater than 1 so here we have considered only the positive sign with 9. Now in the b part we have to find the coordinates of foci and vertices. Now using these results which are given in the key idea the coordinates to the new axis is equal to plus minus ae and capital Y is equal to 0 which is equal to now plus minus ae will be equal to as a square is equal to 81 and e is root 97 by 9 so this is equal to plus minus root 97 now with respect to the whole axis the coordinates of foci will be root is equal to plus minus ae that is plus minus root 97 and y plus 1 is equal to 0 which is equal to plus minus root 97 plus 2 and minus 1. Now the coordinates to the new axis a0 which is equal to as a square is equal to 81 so this is equal to plus minus 9 0 and coordinates of vertices with respect to the whole axis root is equal to plus minus ae which is plus minus 9 and plus 1 is equal to 0 which is equal to plus minus 9 plus 2 and minus 1 which is equal to plus 9 plus 2 is 11 minus 1 and minus 9 this 2 is minus 7 minus 1. Now let us start with the third part in which we have to find the equation of axis and length of axis. Now using these results which are given in the key idea with respect to the new axis capital Y is equal to with respect to the whole axis y plus 1 is equal to 0 which implies y is equal to minus 1 and the length now here we will consider only the positive sign with minus positive so this is equal to this capital X is equal to 0 and with respect to the whole axis equal to 0 which implies X is equal to 2 length that is the length of the conjugate axis is equal to 2 v so here the length is always positive so this will be equal to 2 into 4 which is equal to 8 and we have to find the length of the latest rectum. Now using this results which is given in the key idea the length is equal to 2 b square over a is equal to 2 into b square is 16 over a which is 9 so we will consider the positive sign. So this is the solution of the given question and that is all for this session hope you all have enjoyed the session.