answer:Reaching Eira is not a simple matter, for she resides in a sacred grove that is hidden from the mortal world. The grove is a place of ancient power, where the threads of magic are woven thick and strong. To find it, one must first attune themselves to the rhythms of the wild, and then follow the subtle whispers of the forest. When I wish to seek Eira's counsel, I typically retreat to a secluded spot, where I can quiet my mind and focus my thoughts. I will then call out to her, using a silent language that only the creatures of the wild can understand. The trees, the wind, and the creatures of the forest serve as messengers, carrying my intent to Eira's sacred grove. As I call out to her, I will often take on the form of a bird, such as a raven or an owl, and fly to the edge of the grove. From there, I will follow the subtle energies that emanate from the grove, allowing myself to be drawn into its heart. Once I enter the grove, I am enveloped by a sense of profound peace and tranquility. The trees seem to lean in, as if to listen, and the air is filled with the scent of blooming wildflowers. Eira will then reveal herself, often taking on a form that is both majestic and serene. Our conversations are not like those of mortals, for we communicate through a language of images, emotions, and intuition. Eira will share her wisdom with me, guiding me through the complexities of the world and helping me to unravel the mysteries of the wild. And I, in turn, will share my experiences and insights, that she may better understand the workings of the world beyond the sacred grove.

answer:wanted to show Robin they appreciate the loan

answer:I love to read books. However, I rarely have enough time. Reading books has always been my favorite hobby, but with work and family obligations, I struggle to find time to indulge in it.

answer:Let F_1=(10,25) and F_2=(50,65) be the foci of the ellipse, denoted by mathcal{E}. Suppose mathcal{E} is tangent to the y-axis at some point X. To find the length of the major axis AB, we use the fact that mathcal{E}'s constant sum of distances from any point on the ellipse to the foci, PF_1 + PF_2, equals the length AB. For X, since mathcal{E} is tangent at this point and X lies on the y-axis, we have XF_1 + XF_2 as the smallest possible such distance that also equals k. Consider the reflection of F_1 across the y-axis to the point F_1'=(10,-25). Thus, for a point P on the y-axis PF_1 + PF_2 = PF_1' + PF_2. By the triangle inequality PF_1' + PF_2 geq F_1'F_2 and equality holds if P lies on segment overline{F_1'F_2}. Calculating F_1'F_2: [ F_1'F_2 = sqrt{(50-(-10))^2 + (65-(-25))^2} = sqrt{60^{2} + 90^{2}} = sqrt{3600 + 8100} = sqrt{11700} = 10sqrt{117} ] Thus, the length of the major axis of the ellipse is boxed{10sqrt{117}}.