 Hello and how are you all today? My name is Priyanka and I shall be helping you with the following question. It says in each of the exercise 7 to 15, find the equation of the hyperbola satisfying the given condition. The foci is given to us as plus minus 4 comma 0 and the latest rectum is of length 12. Now here the foci lies on the x-axis, thus the general equation of of hyperbola is x square by e square minus y square by v square is equal to 1. Now here we are given no, we know that foci is equal to plus minus c comma 0 and here foci is given to us as plus minus 4 comma 0 thus we have the value of c as 4. Also we know that the length of the latest rectum is equal to 2 b square by a and here the latest rectum is of length 12. Thus we have 2 b square by a equals to 12. That implies b square is equal to 12 a by 2 that is equal to 6a. Since we know that b is equal to sorry b square is equal to c square minus a square on substituting the values we have b square as 6a is equal to c square will be for the whole square minus a square that further implies a square plus 6a minus 16 is equal to 0. On factorizing the middle term we have a minus 2 and a plus 2 is equal to 0. This is the value of a can be equal to 2 or equal to minus 8. Since a cannot be negative therefore we have the value of a as 2. Now b square was found out to be 6a that means b square is equal to 12. Now we know the value of a and b therefore we can write down the general equation of the hyperbola as square by 4 that is e square minus y square by 12 that is b square is equal to 1. So this is the required answer to the question that was given to us. Hope you understood the whole concept well and have a very nice day ahead.