Find a cartesian equation for the polar curve

Question

given $r= 9 \tan(\theta)\sec(\theta)$

quite clearly we can do simple trig identities and make this much simpler to work with:

Since:

$\tan(\theta)= \frac{\sin(\theta)}{\cos(\theta)}$ and $\sec(\theta)= \frac{1}{\cos(\theta)}$

Using $x=r\cos(\theta)$ and $y=r\sin(\theta)$, we get:

$r=9\frac{\sin(\theta)}{\cos(\theta)} \times \frac{1}{\cos(\theta)}$

Using substitution we get:

$r = 9\frac{\frac{x}{r}}{\frac{x}{r}} \times \frac{r}{x}$

which makes out to be...$1 = \frac{9}{x}$ which is not a parabola

Answer 1

$$r= 9 \tan(\theta)\sec(\theta)=9\frac{\sin \theta}{\cos ^2\theta}$$ multiply by $r$ $$r^2=9r\frac{\sin \theta}{\cos ^2\theta}$$ $$r^2\cos^2\theta=9r\sin\theta$$ $$x^2=9y$$

Answer 2

Variant: $$r=9\frac{\tan\theta}{\cos\theta}\implies r\cos\theta=9\tan\theta\iff x=9\frac yx\implies y=\frac{x^2}9.$$

Answer 3

$$r= 9 \tan(\theta)\sec(\theta) , \quad r=9 \tan(\theta)/\cos(\theta)$$

$$r\cos(\theta)= x = 9 \tan(\theta) = 9 \, y/x $$

$$ 9 y = x^2 $$