Show that the curve has two tangents I'm a little stuck on a math problem that reads as follows:

Show that the curve $x = 5\cos(t), y = 3\sin(t)\, \cos(t)$ has two tangents at $(0, 0)$ and find their equations

What I've Tried


*

*$ \frac{dx}{dt} = -5\sin(t) $

*$ \frac{dy}{dt} = 3\cos^2(t) - 3\sin^2(t) $ because of the product rule. We can simplify this to $ 3(\cos^2(t) - \sin^2(t)) \rightarrow 3\cos(2t) $

*In order to get the slope $ m $, we can write $$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $$

*Solving for $ \frac{dy}{dx} $ as follows:


*

*$ \frac{3\cos(2t)}{-5\sin(t)} $ can be rewritten as

*$ \frac{-3}{5}(\cos(2t)\csc(t)) = m $


*Plugging $ (0,0) $ back into the equations of $ x $ and $ y $ we have as follows:


*

*$ 5\cos(t) = 0 \rightarrow t = \frac{\pi}{2} $


*

*Note: I'm unsure what happens to the $ 5 $


*$ \frac{dx}{dt} $ evaluated at $ t = \frac{\pi}{2} $ gives us $ -5\sin(\frac{\pi}{2}) = -5 $

*$ \frac{dy}{dt} $ evaluated at $ t = \frac{\pi}{2} $ gives us $ 3\cos(\frac{2\pi}{2}) \rightarrow 3\cos(\pi) = -3 $

*$ \frac{dy}{dx} = \frac{3}{5} $


*Continuing on, if we add $ \pi $ to the value of $ t $ we get $ t = \frac{3\pi}{2} $. Plug the new value of $ t $ into the equations of $ x $ and $ y $


*

*$ \frac{dx}{dt} $ evaluated at $ t = \frac{3\pi}{2} $ gives us $ -5sin(\frac{3\pi}{2}) = 5 $

*$ \frac{dy}{dt} $ evaluated at $ t = \frac{3\pi}{2} $ gives us $ 3cos(\frac{6\pi}{2}) \rightarrow 3cos(3\pi) = -3 $

*$ \frac{dy}{dx} = -\frac{3}{5} $


*We now have our two slopes of the tangent lines:


*

*$ y = -\frac{5}{3}x $

*$ y = \frac{5}{3}x $



The issue is that webassign is claiming that the slopes are wrong as can be seen here:

Here is the solution in graph form that is correct:

p.s. My apologies if this is a repost. I've seen this response Show that the curve x = 5 cos t, y = 4 sin t cos t has two tangents at (0, 0) and find their equations. and followed it already with no avail.
 A: The user kccu already found your mistake. Here is a shorter way to do the question. 
The curve is $(5\cos t, 3\sin t\,\cos t)$. With $0\leq t<2\pi$, it goes through the origin two times: when $t=\pi/2$ and when $t=3\pi/2$. Then derivative is, as you say,
$$
(-5\sin t, 3\cos 2t). 
$$
At the two values of $t=\pi/2$ and $t=3\pi/2$ we get 
$$
(-5,-3),\ \ \text{ and }\ \ (5,-3).
$$
The lines are then $(x,y)=t(-5,-3)$ and $(x,y)=t(5,-3)$. That is
$$
y=-\frac35\,x \ \ \ \text{ and } y=\frac35\,x.
$$
So the slopes are $3/5$ and $-3/5$. 
A: You computed the slopes as $\frac{dy}{dx}=\frac{3}{5}$ and $\frac{dy}{dx}=-\frac{3}{5}$, but for some reason when you wrote the equations of the tangent lines you took the reciprocal of these slopes and wrote $y=\frac{5}{3}x$ and $y=-\frac{5}{3}x$. They should be $y=\frac{3}{5}x$ and $y=-\frac{3}{5}x$. Other than that, your method looks correct.
A: You may like to use a short method to prove this.
Eliminate t to get the Cartesian equation of curve as $f(x,y)=9x^4-225x^2+625y^2$
 Now, equate the least degree terms to zero to get tangents at $(0,0)$ which gives $y=3x/5$ and $y=-3x/5$.
