4.0 Parametric Co-ordinates

The equations $$x = a\sec \phi {\text{ and }}y = b\tan \phi $$ together represent the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$ where $\phi $ is a parameter.

Proof: Let $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ be the hyperbola with centre $C$ and transverse axis $AA'$. Therefore, circle drawn with centre $C$ and segment $AA'$ as diameter is called auxiliary circle of the hyperbola.

$\therefore $ Equation of auxiliary circle is $${x^2} + {y^2} = {a^2}$$

Let $P(x,y)$ be any point on the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$. Draw $PN$ perpendicular to $X-$axis.

Let $NQ$ be a tangent to the auxiliary circle ${x^2} + {y^2} = {a^2}$. Join $CQ$ and $\angle QCN = \phi $ such that $\phi $ is the eccentric angle of $P$ $(0 \leqslant \phi \leqslant 2\pi )$.

Since $Q \equiv (a\cos \phi ,a\sin \phi )$, $$x = CN = \frac{{CN}}{{CQ}}.CQ = \sec \phi .a$$$$x = CN = a\sec \phi $$$$\therefore P(x,y) \equiv (a\sec \phi ,y)$$

$\because $ $P$ lies on $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$

$$\begin{equation} \begin{aligned} \frac{{{a^2}{{\sec }^2}\phi }}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \\ \frac{{{y^2}}}{{{b^2}}} = {\sec ^2}\phi - 1 = {\tan ^2}\phi \\ y = \pm b\tan \phi \\ y = b\tan \phi {\text{ (}}P{\text{ lies in first quadrant)}} \\\end{aligned} \end{equation} $$