# How do you convert 2x+y^2 = 5 to polar form?

Nov 13, 2016

#### Explanation:

Substitute $\left(r \cos \left(\theta\right)\right)$ for x and ${r}^{2} {\sin}^{2} \left(\theta\right)$ for ${y}^{2}$:

${r}^{2} {\sin}^{2} \left(\theta\right) +$2rcos(theta) - 5 = 0#

Use the positive root of the quadratic formula:

$r = \frac{- 2 \cos \left(\theta\right) + \sqrt{4 {\cos}^{2} \left(\theta\right) + 20 {\sin}^{2} \left(\theta\right)}}{2 {\sin}^{2} \left(\theta\right)}$

$r = \frac{- \cos \left(\theta\right) + \sqrt{{\cos}^{2} \left(\theta\right) + 5 {\sin}^{2} \left(\theta\right)}}{{\sin}^{2} \left(\theta\right)}$

$r = \frac{- \cos \left(\theta\right) + \sqrt{1 + 4 {\sin}^{2} \left(\theta\right)}}{{\sin}^{2} \left(\theta\right)}$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://socratic.org/questions/how-do-you-convert-2x-y-2-5-to-polar-form", "fetch_time": 1582334305000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2020-10/segments/1581875145621.28/warc/CC-MAIN-20200221233354-20200222023354-00225.warc.gz", "warc_record_offset": 566506608, "warc_record_length": 6063, "token_count": 296, "char_count": 735, "score": 4.28125, "int_score": 4, "crawl": "CC-MAIN-2020-10", "snapshot_type": "longest", "language": "en", "language_score": 0.32328346371650696, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00034-of-00064.parquet"}