Question

Prove that , For and elipse x^2/a^2+y^2/b^2=1, the radius of curvature is a^2b^2/P^3 where P is the length of the perpendicular from the origin upon the tangent at any point​

Answer 1

Step-by-step explanation:

• we know radius is perpendicular to tangentand we know radius is half of diameter
• in the above diagram diameter is perpendicular to tangent , so angle between diameter and tangent is 90 and they are perpendicular to each other