2. Let T, and T2 be two half circles like in Figure 2: they are symmetric relati

Question

2. Let T, and T2 be two half circles like in Figure 2: they are symmetric relative to the t-axis. Take P an arbitrary point on T and let Qbe its reflection about the e axis The hyperbolic line determined by P and Q is the half-circle with center at O you can see in the figure. (i) Show that h(PR) = h(RQ). (This implies h(PQ) = 2h(m) (ii) Determine the point P on the half-circle1 for which h (PQ) is minimal. Hint. Use that h(PR) = ln aca-ota and the fact that csca cota increases if a increases (so the question is: what can you say about the line OP ifthe angle a is maximal?) Figure 2.

Explanation / Answer

given that t1 and t2 are symmetric half circles, that imples P=Q,(this implies equidistant from

x- axis and y-axis).

1) angle from O to P is alpha( a ), the same for Q but 180- alpha( a ). and angle from Q to Ris same alpha( a )

|h(PR)|= ln( 1/ (csc a- cot a) )=|h(RQ)|.

as cosec 180-a= cosec a;

2) given that as alpha (a ) increses the value of h(PR ) decreases. the value will be minimum when the value of alpha is heightest. the maximum value that alpha can take here is 90 degrres .

as alpha (a ) 90 it reaches the point R. then P and R coincides. the value of h(PR) tends to zero(0).

h(PR)= ln(1/ csc a- cot a);

when a= 90,csc a= 1;

when a=90 ,cot a =0;

then ln(1/1-0)= 0;

the height is minimal. by this the half circle t1 coincides with the middle half circle.

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