Let $\alpha ,\beta$, and $\gamma$be nonzero constants. Show that the graph of $r=\frac{\alpha }{\beta +\gamma \cos \theta }$is a conic section with eccentricity $\left|\frac{\gamma }{\beta }\right|$and directrix $x=\frac{\alpha }{\gamma }$.

The non-zero constants $\alpha ,\beta$and $\gamma$of the equation $r=\frac{\alpha }{\beta +\gamma \cos \theta }$.

The equation $r=\frac{\alpha }{\beta +\gamma \cos \theta }$is not in the standard form of the polar equation

$r=\frac{eu}{1+e\cos \theta }$

Convert the equation to the standard form by dividing with $\beta$in the numerator and in the denominator.

Then,

$r=\frac{\frac{\alpha }{\beta }}{\frac{\beta +\gamma \cos \theta }{\beta }}$since dividing by$\beta$

$r=\frac{\frac{\alpha }{\beta }}{\frac{\beta }{\beta }+\frac{\gamma \cos \theta }{\beta }}$

Now multiply and divide the numerator by $\gamma$.then the equation becomes as follows,

$$\begin{gathered} r=\frac{\frac{\gamma }{\gamma }·\frac{\alpha }{\beta }}{1+\frac{\gamma }{\beta }·\cos \theta } \\ r=\frac{\frac{\gamma }{\beta }·\frac{\alpha }{\gamma }}{1+\frac{\gamma }{\beta }·\cos \theta } \end{gathered}$$

The equation $r=\frac{\frac{\gamma }{\beta }·\frac{\alpha }{\gamma }}{1+\frac{\gamma }{\beta }·\cos \theta }$ is in the standard form.

Now compare the equation $r=\frac{\frac{\gamma }{\beta }·\frac{\alpha }{\gamma }}{1+\frac{\gamma }{\beta }·\cos \theta }$with $r=\frac{eu}{1+e\sin \theta }$.

For the polar equation $r=\frac{\frac{\gamma }{\beta }·\frac{\alpha }{\gamma }}{1+\frac{\gamma }{\beta }·\cos \theta }$the eccentricity is equal to $\frac{\gamma }{\beta }$which is taken as Positive

So it is $\left|\frac{\gamma }{\beta }\right|$and the directrix is $y=\frac{\alpha }{\gamma }$.

Therefore the eccentricity is $\left|\frac{\gamma }{\beta }\right|$and the directrix is $y=\frac{\alpha }{\gamma }$.

Hence it is proved.