 Hello and welcome to the session. In this session we will discuss other forms of the hyperbola. Now we know that the standard equation of a hyperbola is x square over a square minus y square over b square is equal to 1. And also we know that the simplicity of this hyperbola is greater than 1 and here a and b are the constants. And this is the horizontal hyperbola. Now let us discuss other forms of hyperbola. First is the vertical hyperbola. Now the equation of the vertical hyperbola is y square over a square minus x square over b square is equal to 1. Now its transverse axis is along the y-axis and its conjugate axis is along the x-axis. And the coordinates of its pokai 0 plus minus ae and the coordinates of the vertices are 0 plus minus a and the equations of the directories are y is equal to plus minus a over e and the eccentricity which is greater than 1 that is e is given by the formula that is square root of a square plus b square whole upon a square. The length of the latest rectum is 2 b square over a then the length of the transverse axis 2 a and its equation is equal to 0 and the length of the conjugate axis is 2 b and its equation is y is equal to 0. So these are all the properties of a vertical hyperbola. So this is a vertical hyperbola. Here f and f1 are the pokai. Of the given hyperbola and the coordinates of the pokai are 0 plus minus ae and a and a1 are the vertices of the given hyperbola. Those coordinates are 0 a and 0 minus a and z and z dash are the directories whose equations are y is equal to a upon e and y is equal to minus a upon e and a a1 is the transverse axis whose equation is x is equal to 0 and whose length is 2 a and b b1 is the conjugate axis whose equation is y is equal to 0 and length is equal to 2 b and l1 and l dash l1 dash are the latest rectum of the given hyperbola whose length is 2 b square upon a. So this is a vertical hyperbola whose equation is y square over a square minus x square over b square is equal to 1. Now let us discuss the conjugate hyperbola. Now if x square over a square minus y square over b square is equal to 1 with a given hyperbola square over a square plus y square over b square is equal to 1 that is y square over b square minus x square over a square is equal to 1 or you can write x square over a square minus y square over b square is equal to minus 1 x conjugate hyperbola. Now here in the part 1 this is a hyperbola whose equation is x square over a square minus y square over b square is equal to 1 with a a1 as its transverse axis and b b1 as its conjugate axis. Now in the second part we have the conjugate hyperbola of the given hyperbola and the equation of this hyperbola that is the conjugate hyperbola of the given hyperbola is x square over a square minus y square over b square is equal to minus 1 and in this case this is a hyperbola whose transverse and conjugate axis are respectively the conjugate and transverse axis of the given hyperbola that is in this case a a1 is the conjugate axis and e b1 is the transverse axis of this hyperbola so this is called the conjugate hyperbola of the given hyperbola. Now let us click this hyperbola as equation number 1 and its conjugate hyperbola as equation number 2. Now if e is the eccentricity of the hyperbola which is given by equation number 1 then b square is equal to a square into e square minus 1 over and f is the eccentricity of the hyperbola that is the conjugate hyperbola which is given by equation number 2 when a square is equal to b square into e square minus 1 1 over e square is equal to 1. Now let us discuss the hyperbola for which the axis are parallel to the axis of coordinates. Now consider the given hyperbola whose axis are parallel to the coordinate axis and the center of the given hyperbola is o dash whose coordinates are h in. Now the equation of the hyperbola a square minus y minus k whole square over b square is equal to this b equation a. Now the equation a reduced square minus m y square plus 2 g x i plus c is equal to 0 where that it represents a hyperbola whose axis are parallel to the axis of coordinates the journal equation of the hyperbola. Now when the coordinate summing directrix of definition that is the distance of n point p is equal to into the distance of the point p from the directrix of the corresponding directrix m plus n is equal to 0. Then by definition we find whole square the distance of point p from the directrix that is modulus of l x plus m y plus m m square the whole into whole square plus y minus beta whole square this whole is equal to e square into l x plus m y plus m. Now this equation reduces to the form a x square plus 2 h x y plus b y square f y plus c is equal to the journal equation that the equation of the hyperbola is a second degree equation and the constants a h b are such as greater than a. Now let us discuss one example. Now here we have to find the equation of the hyperbola whose focus is and the simplicity is root x. Now let us start with the solution. Now let p x y be any point then by definition of hyperbola p from the focus that is now the coordinates of focus are 3 x so the distance of the point p from the focus will be x minus 3 whole square that is square root of x minus 3 whole square plus y minus that is root x into that is the corresponding directrix modulus of 6 x plus 3 y minus 3 whole upon as the equation of the directrix here is given as 6 x plus 3 y minus 3 is equal to 0. Now squaring both sides plus y minus 6 whole square is equal to into 6 x plus 3 y minus 3 whole square that is 36 plus 3 that is 9 p into 2 is 6 and 3 into 15 is 45. Now in cross multiplying this will give 15 into now x minus 3 whole square is x square plus 9 minus 6 x plus y minus 6 whole square is y square plus 36 minus 12 y the whole is equal to 12 plus 9 y square plus 9 plus 36 x y 18 y minus 3 we get plus 72 x y plus 18 x plus 154 y is equal to 0 and here a is equal to 57 that is the coefficient of x square as p into d that is the coefficient of y square therefore that is 171 which is 1296 is greater than 171 equation denoted by capital B hence the equation which is denoted by capital B is the equation of the hyperbola. Session we have learnt about are the forms of hyperbola and journal equation of the hyperbola. This completes our session hope you all have enjoyed the session.