Conics Queston (1 Viewer)

[Sparcod]

For hyperbola,x2/a2-y2/b2=1

P(a sec 0, b tan 0) and Q(a sec (-0), b tan(-0) ) are extremeties of the latus rectum x=ae.

Show that
a) sec 0=e
b)PQ has length 2b^2/a


For a), I tried finding the equation of PQ. I think that's the first step.

 

[haboozin]

For hyperbola,x2/a2-y2/b2=1

P(a sec 0, b tan 0) and Q(a sec (-0), b tan(-0) ) are extremeties of the latus rectum x=ae.

Show that
a) sec 0=e
b)PQ has length 2b^2/a


For a), I tried finding the equation of PQ. I think that's the first step.

a. extremeties of the latus rectum

so asec0 = ae (since by definition the latus rectum is the perpendicular line throught he focus)

so sec0 = e

as req'd

b. again the top half and bottom half is symetrical

so 2 x the top half
and the top half is y coordinates of p

so 2 x b tan0 (dont use 0 again it looks liek ZERO DEGREES use @)

but b^2 = a^2 (e^2 - 1)

by definition - I HOPE YOU KNOW THIS.

next

tan^2@ = sec^2@-1

anotehr identity

since sin^2@ + cos^2@ = 1
and divide by cos^2@

(again u should know this)

apply them both to 2 x b tan0
and u get

2b^2/a

as req'd

 

[_ShiFTy_]

a) is actually quite simple

since " P(a sec 0, b tan 0) and Q(a sec (-0), b tan(-0) ) are extremeties of the latus rectum x=ae"

Then Xp = ae
asec0 = ae
sec0=e

 

[Sparcod]

Nice stuff boys.
Thanks very much.