# How do you find the points of horizontal tangency of r=2csctheta+5?

Jan 19, 2018

Points of horizontal tendency are $\theta = \frac{\pi}{2}$ and $\theta = \frac{3 \pi}{2}$

#### Explanation:

Let us convert $r = 2 \csc \theta + 5$ to Cartesian coordinates by using $r \cos \theta = x$, $r \sin \theta = y$ and ${r}^{2} = {x}^{2} + {y}^{2}$. We can write $r = 2 \csc \theta + 5$ as

$r \sin \theta = 2 + 5 \sin \theta$ or $y = 2 + \frac{y}{\sqrt{{x}^{2} + {y}^{2}}}$

i.e. $\frac{y}{\sqrt{{x}^{2} + {y}^{2}}} = y - 2$

or ${y}^{2} / \left({x}^{2} + {y}^{2}\right) = {\left(y - 2\right)}^{2}$

or $\left({x}^{2} + {y}^{2}\right) {\left(y - 2\right)}^{2} = {y}^{2}$

Now horizontal tangents are there where $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ and hence finding $\frac{\mathrm{dy}}{\mathrm{dx}}$ using implicit differentiiation, we get

$2 \left({x}^{2} + {y}^{2}\right) \left(y - 2\right) \frac{\mathrm{dy}}{\mathrm{dx}} + {\left(y - 2\right)}^{2} \left(2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

or $\frac{\mathrm{dy}}{\mathrm{dx}} \left[2 \left({x}^{2} + {y}^{2}\right) \left(y - 2\right) + 2 y {\left(y - 2\right)}^{2} - 2 y\right] = - 2 x {\left(y - 2\right)}^{2}$

and hence $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ when either $y = 2$ or $x = 0$

Now $y = 2$ means $r \sin \theta = 2$ but as equation can be written as $r \sin \theta = 2 + 5 \sin \theta$, this means when $y = 2$, we have $\sin \theta = 0$ or $\theta = 0$. But as function $r = 2 \csc \theta + 5$ s not defined at $\theta = 0$, we ignore this.

$x = 0$ means $r \cos \theta = 0$ or $\cos \theta = 0$, i.e. $\theta = \frac{\pi}{2}$ or $\frac{3 \pi}{2}$ and hence points of horizontal tendency are $\theta = \frac{\pi}{2}$ and $\frac{3 \pi}{2}$

graph{(x^2+y^2)(y-2)^2=y^2 [-10, 10, -5, 5]}