Lengths of polar subtangent and polar subnormal

Draw a line NT through the origin perpendicular to the radius vector of the point P on the curve. If PT is the tangent and PN the normal to the curve at P, then^6.8

Figure 6.10: The polar subtangent and polar subnormal.

OT = length of polar subtangent,

and

ON = length of polar subnormal

of the curve at P.

In the triangle OPT, $\tan\, \psi = \frac{OT}{\rho}$. Therefore

$$OT = \rho \tan \psi = \rho^2 \frac{d\theta}{d\rho} =\ {\rm length\ of\ polar\ subtangent}.$$ (6.16)

In the triangle OPN, $\tan \psi = \frac{\rho}{ON}$. Therefore

$$ON = \frac{\rho}{\tan \psi} = \frac{d\rho}{d\theta} =\ {\rm length\ of\ polar\ subnormal}.$$ (6.17)

The length of the polar tangent (= PT) and the length of the polar normal (= PN) may be found from the figure, each being the hypotenuse of a right triangle.

Example 6.8.1   Find lengths of polar subtangent and subnormal to the lemniscate $\rho^2 = a^2\cos\, 2\theta$.

Solution. Differentiating the equation of the curve as an implicit function with respect to $\theta$, or, $2 \rho \frac{d\rho}{d\theta} =\ - 2 a^2 \sin 2 \theta$, $\frac{d\rho}{d\theta} =\ -\frac{a^2 \sin 2\theta}{\rho}$.

Substituting in (6.16) and (6.17), we get

length of polar subtangent = $- \frac{\rho^3}{a^2 \sin 2\theta}$,
length of polar subnormal = $- \frac{a^2 \sin 2\theta}{\rho}$.

If we wish to express the results in terms of $\theta$, find $\rho$ in terms of $\theta$ from the given equation and substitute. Thus, in the above, $\rho = \pm a \sqrt{\cos 2 \theta}$; therefore

length of polar subtangent = $\pm a \cot 2 \theta \sqrt{\cos 2 \theta}$.