Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 11

In plane polar coordinates, $r$ is distance from the origin and $\theta$ is the angle from positive $x$ toward positive $y$. The conversion to Cartesian coordinates isConversely $r=\sqrt{x^2+y^2}$, with the angular coordinate undefined at the origin.

Multiplying the polar equation by its denominator gives $Ar-Bx=1$. Squaring and using $r^2=x^2+y^2$ yieldsTake $C=\sqrt{A^2-B^2}>0$. Completing the square givesThe standard ellipse equation is consequentlywhose centre is $\boxed{(B/C^2,0)}$. Squaring has introduced no unwanted branch: on this ellipse the minimum of $1+Bx$ is $A/(A+B)>0$, so $Ar=1+Bx$ follows with the correct positive sign.

Read the squared semiaxis lengths from the denominators in the standard equation obtained in (b). With the positive choice of $C$ there,Since $C^2=A^2-B^2<A^2$, the horizontal semiaxis $a$ is the larger. Thus these are the semi-major and semi-minor axes, respectively.