Annoying COnics Q (1 Viewer)

[richz]

ok this is from cambridge ex 3.2 q 8b

& - thi

q: P(sec@, btan@) and Q(acos&, bsin&) lie on the ellipse x^2/a^2 +y^2/b^2 = 1

B) If PQ subtends a right angle at (a,0) show than tan(@/2)tan(&/2) = -b^2/a^2

heres my working so far:

M pa (btan@/a(sec@-1))
M qa (btan&/a(sec&-1))

(btan@)/(a(sec@-1)) * (btan&)/(a(sec&-1)) = -1
(b^2/a^2) * (sin@sin&)/(cos@cos&) * 4sin(&/2)sin@/2

where now??

thnx in advance

 

[FinalFantasy]

ok this is from cambridge ex 3.2 q 8b

& - thi

q: P(sec@, btan@) and Q(acos&, bsin&) lie on the ellipse x^2/a^2 +y^2/b^2 = 1

B) If PQ subtends a right angle at (a,0) show than tan(@/2)tan(&/2) = -b^2/a^2

heres my working so far:

M pa (btan@/a(sec@-1))
M qa (btan&/a(sec&-1))

(btan@)/(a(sec@-1)) * (btan&)/(a(sec&-1)) = -1
(b^2/a^2) * (sin@sin&)/(cos@cos&) * 4sin(&/2)sin@/2

where now??

thnx in advance

"M pa (btan@/a(sec@-1))
M qa (btan&/a(sec&-1))"
M pa=bsin@\a(1-cos@)=bcos(@\2)\asin(@\2)
M qa=bcos(&\2)\asin(&\2)
bcos(@\2)\asin(@\2)*bcos(&\2)\asin(&\2)=-1
b²\a² cos(@\2)cos(&\2)=-sin(@\2)sin(&\2)
tan(@\2)tan(&\2)=-b²\a²

just make da stuff into sin and cos den use half angles