Question

Question #d9987

Answer 1

`-10rsin(\theta) + 25 = r^2cos^2(\theta)`

Explanation:

To convert any Cartesian equation to polar form you just need to know that `x = rcos(\theta)` and `y = rsin(\theta)`.

So in this equation we can replace all the x's and y's with the above substitutions to get:

`-10rsin(\theta) + 25 = r^2cos^2(\theta)`

This is an equation in polar form.

Answer 2

Substitute `rsin(theta)` for y and `r^2cos^2(theta)` for `x^2`
Write the resulting quadratic in standard form.
Use the quadratic formula to obtain `r` as a function of `theta`

Explanation:

Given: `-10y+25=x^2`

Here is a graph of the original equation:

Substitute `rsin(theta)` for y and `r^2cos^2(theta)` for `x^2`

`-10sin(theta)r+25=cos^2(theta)r^2`

Write in standard form:

`0=cos^2(theta)r^2+10sin(theta)r-25`

Use the quadratic formula:

`r = (-10sin(theta)+sqrt((10sin(theta))^2-(4)(cos^2(theta))(-25)))/(2cos^2(theta))`

`r = 5(1-sin(theta))/cos^2(theta)`

Here is a graph of the polar equation:

It looks like the graphing calculator is not handling the `lim_(theta to pi/2) 5(1-sin(theta))/cos^2(theta)`, correctly, but the graphs are the same, except for that glitch.