1.

E1:x2a2+y2b2−1=0,(a>b) and E2:x2k2+y2b2−1=0,(k<b) is inscribed in E1. If E1 and E2 have same eccentricities and length of minor axis of E2=p×LLR of E1, then p=

Answer» E1:x2a2+y2b21=0,(a>b) and E2:x2k2+y2b21=0,(k<b) is inscribed in E1. If E1 and E2 have same eccentricities and length of minor axis of E2=p×LLR of E1, then p=


Discussion

No Comment Found