Precalculus math hw help?!

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1 Answer
May 18, 2018

The answer is #[B]#.

Explanation:

Given #x=-tant# and #y=-sect#, where #-pi/2<=t<=pi/2#

at #t=0#, #x=0# and #y=-1# i.e. point #(0,-1)# lies on curve;

as #t->-pi/2# (from RHS), #x->oo# and #y->-oo# i.e. point #(oo,-oo)# and as #t# moves from #-pi/2# to #0#, point moves from #(oo,-oo)# to #(0,-1)# i.e. from #Q4# to to #(0,-1)# on #y#-axis

and as #t->pi/2# (from LHS), #x->-oo# and #y->-oo# i.e. point #(-oo,-oo)# and as #t# moves from #0# to #pi/2#, point moves from #(0,-1)# to #(-oo,-oo)# i.e. from #(0,-1)# on #y#-axis to #(-oo,-oo# in #Q3#

Hence answer is #[B]#.