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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/15/d/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 15 / d / Solution by

Codex  0 2026-10-18

Put

c = a^{2} + b^{2}

and denote the distances to

(- c, 0)

and

(c, 0)

by

d_{-}

and

d_{+}

. Direct expansion using

c^{2} = a^{2} + b^{2}

gives

d_{-}^{2} d_{+}^{2} = (a cosh t + c)^{2} + b^{2} sinh^{2} t = (c cosh t + a)^{2}, = (a cosh t - c)^{2} + b^{2} sinh^{2} t = (c cosh t - a)^{2} .

Since

c > a > 0

and

cosh t \geq 1

, both expressions inside the final square roots are positive. Hence

d_{-} - d_{+} = 2 a

for every point of this right branch. This is the focal-distance identity for a hyperbola; on the left branch the signed difference would reverse, while its absolute value would remain

2 a

.

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