Q. 16 - Calculus

Question

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 $x=\frac{\alpha }{\gamma }$.

Step-by-Step Solution

Answer

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

Step 1. Given information:

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

Step 2. Converting the equation to the standard form:

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 $\beta$

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

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

Step 3. multiplying and dividing the numerator:

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.

Step 4. Comparing the equation for finding the eccentricity:

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