1. The problem statement, all variables and given/known data Find an equation in standard form for the conic , a)Ellipse: vertices (0,0) & (0,8), foci (0,2) & (0,6). b) hyperbola: vertices (-3,0) & (3,0), asymptotes y= ±2x 2. Relevant equations Ellipse: (x-h)^2/b^2+(y-k)^2/a^2=1, vertices ± a from center, foci ± c from center, c^2=a^2-b^2, a<b, ceneter = (h,k) Hyperbola: (x-h)^2/a^2-(y-k)^2/b^2=1, vertices ± a from center, foci ± c from center, c^2=a^2+b^2, ceneter = (h,k), asymptotes= k±b/a(x-h) 3. The attempt at a solution a) foci are ±c from center, the foci are (0,2) & (0,6) thus h must = 0, c must= 2, c^2=4, and k= 4 I know c, and the center (0,4) vertices are ± a from the center thus if the center is (0,4) when vertices are ( 0,0) & (0,8), a must= 4 and a^2=16 since I know have a^2 and c^2 I can solve for b^2 4=16-b^2, b^2=12 I can now solve the problem and write the equation in the form of (x-h)^2/b^2+(y-k)^2/a^2=1. x^2/16+(y-4)^2/16=1 End part A b) vertices are ± a from center @ (±3,0) thus the center is (0,0) h=0, k=0 and a=3 and a^2=9 asymptotes are in the form of k±b/a(x-h) thus 0 ± b/3(x-0)= ±2x, b must =6 b^2=36 I can now write the equation in the form of (x-h)^2/a^2-(y-k)^2/b^2=1 as: x^2/9-y^2/36=1 End part B That is all my work/solutions for this problem. Any help with what I may have done wrong and how to learn how to correct is, or even just saying "correct" is appreciated thank you.