If A_E is the area of an ellipse with an eccentricity of e= 7 25 and A_F is the area of the shape bounded by the set of points for which two tangents of that ellipse meet at a right angles, then A_E A_F = p q , where p and q are positive co-prime integers. Find (p+q) .

Step-by-step solution

Since e= 7 25 = c a , c=7 k and a=25 k and since a^2-b^2=c^2 in an ellipse, b=24 k . The area of an ellipse is A= a b , so A_E=600 k^2 . The set of points for which two tangents of any curve meet at a right angle is an orthopetic, and an orthopetic for any ellipse is a circle with a radius of a^2+b^2 , and therefore an area of A= (a^2+b^2 ) , so A_F=1201 p k^2 . Therefore, A_E A_F = 600 k^2 1201 k^3 = 600 1201 , so p=600, q=1201 , and p+q=1801