=1 So, a2=169,b2=144 ‌⇒‌‌a=13,b=12 ‌∴‌‌c2=a2−b2⇒169−144=25 ‌⇒‌‌c=5 So, the foci are at (±c,0). So, S=(5,0) and S′=(−5,0) Since, the point B is one end of the minor axis lying on the positive Y-axis. So, B=(0,±b)=(0,12) ∴ The vertices of the triangle are S(5,10),S′(−5,0) and B(0,12) SS′‌=√(−5−5)2+(0−0)2 ‌=√(−10)2=10 SB‌=√(0−5)2+(12−0)2 ‌=√25+144=√169=13 S′B‌=√(0−(−5))2+(12−0)2 ‌=√25+144=√169=13 Since, SB=S′B, So SBS′ is an isosceles triangle. Now, the incenter (Ix,Iy) of a triangle with vertices (x1,y1),(x2,y2),(x3,y3) and opposite side lengths a,b and c is Ix=‌
ax1+bx2+cx3
a+b+c
,Iy=‌
ay1+by2+cy3
a+b+c
Here, (x1,y1)=S(5,0),(x2,y2)=S′(−5,0),(x3,y3)=B(0,12) and a=S′B=13,b=SB=13,C=SS′=10 ∴‌Ix=‌