91Ó°ÊÓ

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

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

$r=\frac{eu}{1+e\sin \mathrm{θ}}$

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

Then,

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

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

$r=\frac{\frac{\mathrm{Î\pm }}{\mathrm{β}}}{1+\frac{\mathrm{γ}}{\mathrm{β}}·\sin \mathrm{θ}}$

Now multiply and divide by $\mathrm{γ}$in the numerator to make it in the standard form.

$r=\frac{\frac{\mathrm{γ}}{\mathrm{γ}}·\frac{\mathrm{Î\pm }}{\mathrm{β}}}{1+\frac{\mathrm{γ}}{\mathrm{β}}·\sin \mathrm{θ}}$

$r=\frac{\frac{\mathrm{γ}}{\mathrm{β}}·\frac{\mathrm{Î\pm }}{\mathrm{γ}}}{1+\frac{\mathrm{γ}}{\mathrm{β}}·\sin \mathrm{θ}}$

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

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

For the polar equation $r=\frac{\frac{\mathrm{γ}}{\mathrm{β}}·\frac{\mathrm{Î\pm }}{\mathrm{γ}}}{1+\frac{\mathrm{γ}}{\mathrm{β}}·\sin \mathrm{θ}}$the eccentricity is equals to $\frac{\mathrm{γ}}{\mathrm{β}}$which is taken as positive .so eccentricity is $\left|\frac{\mathrm{γ}}{\mathrm{β}}\right|$and the directrix is $y=\frac{\mathrm{Î\pm }}{\mathrm{γ}}$.

Hence it is proved.