yet another conics question (1 Viewer)

[currysauce]

P (a sec0 , b tan0) lies on the hyperbola (normal hyper. eqn). with foci S (ae,0) and S' (-ae, 0).

a) show that PS = a(e sec0 - 1) and PS' = a(esec0 +1)
b) Deduce that |PS-PS'| = 2a

i seriously know whats happenig here, but i just can't get it....

anyway thanks

 

[Ogden_Nash]

a) Keep trying with the distance formula. Play around with the b² = a²(e² - 1) and tan²@ + 1 = sec²@ identities and you'll get it eventually.

b) No trouble once you've done a).

 

[:: ck ::]

mm no need for distance formula

remmeber PS/PD = e
Ps = e* PD

u kno the PD is just horizontal distance asec@ - a/e

so PS = e(asec@-a/e)
= a(1-esec@)

cbf doin rest im so over hsc rofl .. u shud bable to do it from there

 

[currysauce]

kool

also could anyone attempt,

9b, pg92, cambridge

thanks if possible