00:01
In this question, we have to show that each point p of the hyperbola, the tangent line bisects the angle between the focal radiae f1p and f2 p.
00:17
Now let hyperbola is x squared divided by a square minus y squared divided by b squared equals to 1.
00:33
Now we know that equation of tangent of hyperbola at point 5 is x by a multiplied by secant 5 minus y by b times y by b tangent 5 equals to 1.
00:58
Now let its gradient is m1 then m1 equals to minus coefficient of x which is minus 1 by a secant 5 divided by coefficient of y which is minus 1 by b tangent phi in a simplified form we can write it as b second 5 divided by a tangent 5 this we can write it as b by a times we can write secant phi as 1 by cosine phi and tangent 5 as sine 5 as sine 5 so it will be cosine 5 divided by sine pi.
02:11
In simplified form, we can write it as m1 equals to b by a times 1 by sine pi.
02:24
Now we know that the gradient of a line which passes through points x1 by 1 and x2 y2 is m equals to y2 minus y1 divided by x2 minus x1.
02:40
So gradient of flying f1p is m2 then m2 equals to b tangent phi minus 0 divided by a seccent phi minus ae.
03:13
So we can write this b tangent 5 divided by we will take common a secund phi minus e.
03:29
Now let represent it by b.
03:41
Now because here x1 y1 is ae.
03:45
0 and x2.
03:47
2 .2 is a secant 5 and b tangent 5 and gradient of line f2 p is m3, then m3 equals to b tangent 5 minus 0 divided by a secend 5 minus bracket minus ae.
04:18
In simplified form you can write it as b tangent 5 divided by a times secant 5 plus c let mark it as c.
04:40
Now because here x1 y1 is minus ae comma 0 and x2 y2 is is a secant d and b tangent b let tangent at p divided the angle between f1 p and f2 p is is alpha and beta...