Question

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

Short answer

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

Step-by-step solution

Step 1. Given information: 

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

Step 2. Converting the equation to the standard form:

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

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

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

Then,

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

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

$r=\frac{\frac{\alpha }{\beta }}{1+\frac{\gamma }{\beta }·\sin \theta }$

Step 3. multiplying and dividing the numerator:

Now multiply and divide by $\gamma$in the numerator to make it in the standard form.

$r=\frac{\frac{\gamma }{\gamma }·\frac{\alpha }{\beta }}{1+\frac{\gamma }{\beta }·\sin \theta }$

$r=\frac{\frac{\gamma }{\beta }·\frac{\alpha }{\gamma }}{1+\frac{\gamma }{\beta }·\sin \theta }$

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

Step 4. Comparing the equation for finding the eccentricity: 

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

For the polar equation $r=\frac{\frac{\gamma }{\beta }·\frac{\alpha }{\gamma }}{1+\frac{\gamma }{\beta }·\sin \theta }$the eccentricity is equals to $\frac{\gamma }{\beta }$which is taken as positive .so eccentricity is $\left|\frac{\gamma }{\beta }\right|$and the directrix is $y=\frac{\alpha }{\gamma }$.

Hence it is proved.