 Hello and welcome to this session. In this session we will discuss a question which says that find the length of transverse and contribute axis, a simplicity, the range of foci and vertices, length of filter spectrum, equations of direct access or the given hyperbola. 4x squared minus 9y squared is equal to minus 36. Now before starting the solution of this question we should know some results. First is if the equation of the hyperbola is given as x squared over a squared minus y squared over b squared is equal to 1 then minus x squared over a squared plus y squared over v squared is equal to 1 that is y squared over b squared minus x squared over a squared is equal to 1 or you can write x squared over a squared minus y squared over b squared is equal to minus 1 conjugate hyperbola where a and b are the constants. Secondly for the conjugate hyperbola the coordinates of the vertices are given as 0, b and 0 minus b and the coordinates of the foci are given as 0 plus minus b e where e is greater than 1 as for the hyperbola a simplicity is greater than 1 then length of the transverse axis is to be the length of the conjugate axis then the length of labor spectrum is 2 a squared over b and the equations as y is equal to plus minus b over e. Also the simplicity e is equal to square root of b squared plus a squared whole upon. Now these results will work out as a key idea for solving out this question. And now we will start with the solution. Now in the question for the given hyperbola we have to find the length of the transverse and conjugate axis then the eccentricity then coordinates of foci vertices length of the labor spectrum and equations of the directorial equation of the given hyperbola is minus 9 y squared is equal to minus 36 and this can be where over 36 minus 9 y squared over 36 is equal to minus 1. Which implies x squared over 9 minus y squared over 4 is equal to minus 1. Now this is the equation of the conjugate hyperbola of the given hyperbola. This is the form of the equation of conjugate hyperbola. Now here l is equal to 9 and l is equal to first of all we will find the length of transverse and conjugate axis for this given hyperbola. Now using these results which are given in the key idea, given length now here b squared is equal to 4 so b will be equal to plus minus 2. Now we know that length is always positive so we will consider the positive value of b into 2 which is equal to 4. Now the length which will be equal to 2 into a squared is 9 so which is equal to now we can find out the eccentricity by using this formula. Now the eccentricity e of the given hyperbola is equal to square root of 9 and b squared is 4 which is equal to square root of plus 9, where the root which is equal to root 13 by 2. Now we know that eccentricity of the hyperbola is greater than 1 so here we are considering only the positive sign with 2. Now by using these results we will find the coordinates of vertices and foci. So this is equal to 0 and b squared is 4 where it will be plus minus 2 into root 13 by 2. So this is equal to 0 plus minus therefore the coordinates 0 plus minus root 30. Now the coordinates what is this? 0 plus minus 3 which is equal to 0. Now b squared is 4 so this will be plus minus so the coordinates of vertices are 0 plus minus 2 and the length of the vector sector by using this formula. Now the length of the vector sector is equal to 2 a squared by b which is equal to 2 into a squared which is 9 now whereas as length is positive and we are having b squared that means b is equal to plus minus 2 so we will consider only the positive sign as length is positive. So there b is 2 therefore the length of the vector sector will be equal to now. And now we will find the equations of the rectices by using this result the equations are drawn by y t is root 13 by 2 so the equations of the rectices will be y is equal to plus minus y 2 and the transverse axis is equal to 4. Length of conjugate axis or the given hyperbola is equal to c is equal to root 13 by 2 and coordinates of vertices are 0 length of the vector sector is equal to 9 and the equations of the rectices are y is equal to root of the given equation and that's all for this session hope you all have enjoyed the session.