 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2/11/b/solution

Multiplying the polar equation by its denominator gives

A r - B x = 1

. Squaring and using

r^{2} = x^{2} + y^{2}

yields

(A^{2} - B^{2}) x^{2} + A^{2} y^{2} - 2 B x = 1.

Take

C = A^{2} - B^{2} > 0

C^{2} (x - \frac{B}{C ^{2}})^{2} + A^{2} y^{2} = 1 + \frac{B ^{2}}{C ^{2}} = \frac{A ^{2}}{C ^{2}} .

The standard ellipse equation is consequently

\frac{( x - B / C ^{2} ) ^{2}}{A ^{2} / C ^{4}} + \frac{y ^{2}}{1/ C ^{2}} = 1,

whose centre is

(B / C^{2}, 0)

. Squaring has introduced no unwanted branch: on this ellipse the minimum of

1 + B x

A / (A + B) > 0

, so

A r = 1 + B x

follows with the correct positive sign.

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