The Orthoptic Nerve of Some Ellipses

Calculus Level 3

If A E A_E is the area of an ellipse with an eccentricity of e = 7 25 e = \frac{7}{25} and A F 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 angle, then A E A F = p q \frac{A_E}{A_F} = \frac{p}{q} , where p p and q q are positive co-prime integers. Find p + q p + q .


The answer is 1801.

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1 solution

David Vreken
Apr 6, 2019

Since e = 7 25 = c a e = \frac{7}{25} = \frac{c}{a} , c = 7 k c = 7k and a = 25 k a = 25k , and since a 2 b 2 = c 2 a^2 - b^2 = c^2 in an ellipse, b = 24 k b = 24k .

The area of an ellipse is A = π a b A = \pi ab , so A E = 600 π k 2 A_E = 600 \pi k^2 .

The set of points for which two tangents of any curve meet at a right angle is an orthoptic , and an orthoptic for any ellipse is a circle with a radius of a 2 + b 2 \sqrt{a^2 + b^2} , and therefore an area of A = π ( a 2 + b 2 ) A = \pi (a^2 + b^2) , so A F = 1201 π k 2 A_F = 1201 \pi k^2 .

Therefore, A E A F = 600 π k 2 1201 π k 2 = 600 1201 \frac{A_E}{A_F} = \frac{600 \pi k^2}{1201 \pi k^2} = \frac{600}{1201} , so p = 600 p = 600 , q = 1201 q = 1201 , and p + q = 1801 p + q = \boxed{1801} .

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