91Ó°ÊÓ

The non-zero constants $\mathrm{Î\pm },\mathrm{β}$and $\mathrm{γ}$of the equation $r=\frac{\mathrm{Î\pm }}{\mathrm{β}+\mathrm{γ}\cos \mathrm{θ}}$.

The equation $r=\frac{\mathrm{Î\pm }}{\mathrm{β}+\mathrm{γ}\cos \mathrm{θ}}$is not in the standard form of the polar equation

$r=\frac{eu}{1+e\cos \mathrm{θ}}$

Convert the equation to the standard form by dividing with $\mathrm{β}$in the numerator and in the denominator.

Then,

$r=\frac{\frac{\mathrm{Î\pm }}{\mathrm{β}}}{\frac{\mathrm{β}+\mathrm{γ}\cos \mathrm{θ}}{\mathrm{β}}}$since dividing by$\mathrm{β}$

$r=\frac{\frac{\mathrm{Î\pm }}{\mathrm{β}}}{\frac{\mathrm{β}}{\mathrm{β}}+\frac{\mathrm{γ}\cos \mathrm{θ}}{\mathrm{β}}}$

Now multiply and divide the numerator by $\mathrm{γ}$.then the equation becomes as follows,

$$\begin{gathered} r=\frac{\frac{\mathrm{γ}}{\mathrm{γ}}·\frac{\mathrm{Î\pm }}{\mathrm{β}}}{1+\frac{\mathrm{γ}}{\mathrm{β}}·\cos \mathrm{θ}} \\ r=\frac{\frac{\mathrm{γ}}{\mathrm{β}}·\frac{\mathrm{Î\pm }}{\mathrm{γ}}}{1+\frac{\mathrm{γ}}{\mathrm{β}}·\cos \mathrm{θ}} \end{gathered}$$

The equation $r=\frac{\frac{\mathrm{γ}}{\mathrm{β}}·\frac{\mathrm{Î\pm }}{\mathrm{γ}}}{1+\frac{\mathrm{γ}}{\mathrm{β}}·\cos \mathrm{θ}}$ is in the standard form.

Now compare the equation $r=\frac{\frac{\mathrm{γ}}{\mathrm{β}}·\frac{\mathrm{Î\pm }}{\mathrm{γ}}}{1+\frac{\mathrm{γ}}{\mathrm{β}}·\cos \mathrm{θ}}$with $r=\frac{eu}{1+e\sin \mathrm{θ}}$.

For the polar equation $r=\frac{\frac{\mathrm{γ}}{\mathrm{β}}·\frac{\mathrm{Î\pm }}{\mathrm{γ}}}{1+\frac{\mathrm{γ}}{\mathrm{β}}·\cos \mathrm{θ}}$the eccentricity is equal to $\frac{\mathrm{γ}}{\mathrm{β}}$which is taken as Positive

So it is $\left|\frac{\mathrm{γ}}{\mathrm{β}}\right|$and the directrix is $y=\frac{\mathrm{Î\pm }}{\mathrm{γ}}$.

Therefore the eccentricity is $\left|\frac{\mathrm{γ}}{\mathrm{β}}\right|$and the directrix is $y=\frac{\mathrm{Î\pm }}{\mathrm{γ}}$.

Hence it is proved.