# Thread: vertical and horizontal tangency to curve

1. ## vertical and horizontal tangency to curve

x=4cos squared (theta)
y=2sin (theta)

after eliminating parameter i got
x=4-y squared

how to find tangents after finding derivitive?

2. So, just to reiterate. You've got

$\displaystyle x=4\cos^{2}(\theta)$ and
$\displaystyle y=2\sin(\theta).$

You want to know when you have vertical and horizontal tangents, correct? If so, I'm not sure I would eliminate the parameter. The reason? Because

$\displaystyle \displaystyle{\frac{dy}{dx}=\frac{\frac{dy}{d\thet a}}{\frac{dx}{d\theta}}}.$

The derivatives on the RHS there are easy to compute, because you have the formulas right there. In addition, by inspection, you can tell that you'll get horizontal tangents precisely when

$\displaystyle \displaystyle{\frac{dy}{d\theta}=0}$ and

$\displaystyle \displaystyle{\frac{dx}{d\theta}\not=0}.$ If they're both equal to zero at a particular value of $\displaystyle \theta$, you may need to do more work. Also, you'll get vertical tangents when

$\displaystyle \displaystyle{\frac{dx}{d\theta}=0}$ and

$\displaystyle \displaystyle{\frac{dy}{d\theta}\not=0}.$

Does this make sense?

3. But since you have eliminated the parameter, should be easy to see that the graph is a parabola with horizontal axis and vertex at (4, 0). It has a vertical tangent at (4, 0) and no horizontal tangent.

Or you could just go ahead and differentiate: if $\displaystyle x= 4- y^2$ then $\displaystyle 1= -2y\frac{dy}{dx}$.

$\displaystyle \frac{dy}{dx}= -\frac{1}{2y}$ is never 0 and does not exist when y= 0.