Ok a question states:
P(acos@,bsin@) lies on ellipse (std eqn.). THe tangnet at P cuts the x-axis at X and the y-axis at Y.
Show that PX/PY = tan2@ and deduce that if P is an extermity of a latus rectum, then PX/PY = (1-e2)/e2.
Ok i did part one no problems, but i would just like to see if the method i chose to prove part two is correct.
If P is an extremity of a latus rectum then, x = ae.
acos@ = ae
cos@ = e
now,
***/|
**/*|
*/**|
/@ _|
This triangle has a base of e, hypotenuse of 1 and a height of 'h'
12 = e2 + h2
therefore h = sqrt(1-e2)
tan@ = sqrt(1-e2) / e
tan2@ = 1-e2 / e2
hence, PX/PY = tan2@ = 1-e2 / e2.
Is this a valid proof using that triangle crap. Because i remember last time i did it another way which was much longer.
Thanks for any clarifications.
P(acos@,bsin@) lies on ellipse (std eqn.). THe tangnet at P cuts the x-axis at X and the y-axis at Y.
Show that PX/PY = tan2@ and deduce that if P is an extermity of a latus rectum, then PX/PY = (1-e2)/e2.
Ok i did part one no problems, but i would just like to see if the method i chose to prove part two is correct.
If P is an extremity of a latus rectum then, x = ae.
acos@ = ae
cos@ = e
now,
***/|
**/*|
*/**|
/@ _|
This triangle has a base of e, hypotenuse of 1 and a height of 'h'
12 = e2 + h2
therefore h = sqrt(1-e2)
tan@ = sqrt(1-e2) / e
tan2@ = 1-e2 / e2
hence, PX/PY = tan2@ = 1-e2 / e2.
Is this a valid proof using that triangle crap. Because i remember last time i did it another way which was much longer.
Thanks for any clarifications.