Answer to Question #111156 in Analytic Geometry for Amoah Henry

Question

Find the Cartesian equations of the lines or curves whose polar equations are (a) r = a (b) r = a cos 2θ
(c) r2 = a2 sin 2θ (e) r = a sec(θ − α)

Answer 1

Note: $r=\sqrt{x^2+y^2}; sin(\theta)=\frac{y}{\sqrt{x^2+y^2}}; cos(\theta)=\frac{x}{\sqrt{x^2+y^2}}$

a) $r=a$

$\sqrt{x^2+y^2}=a\\ x^2+y^2=a^2$

b)$r=acos(2\theta)$

$r=a(2cos^2(\theta)-1)\\ \sqrt{x^2+y^2}=a(2(\frac{x}{\sqrt{x^2+y^2}})^2-1)\\ \sqrt{x^2+y^2}=a(\frac{2x^2}{x^2+y^2}-1)\\ \sqrt{x^2+y^2}=a\frac{x^2-y^2}{x^2+y^2}$

$x^2+y^2=a^2\frac{(x^2-y^2)^2}{(x^2+y^2)^2}\\ (x^2+y^2)^3=a^2(x^2-y^2)^2$

c) $r^2=a^2sin(2\theta)$

$r^2=2a^2sin(\theta)cos(\theta)\\ x^2+y^2=2a^2\frac{y}{\sqrt{x^2+y^2}}\frac{x}{\sqrt{x^2+y^2}}\\ x^2+y^2=2a^2\frac{xy}{x^2+y^2}\\ (x^2+y^2)^2=2a^2xy$

e)$r=asec(\theta-\alpha)$

$r=\frac{a}{cos(\theta-\alpha)}\\ rcos(\theta-\alpha)=a$

$r(cos(\theta)cos(\alpha)+sin(\theta)sin(\alpha))=a$

$\sqrt{x^2+y^2}(\frac{x}{\sqrt{x^2+y^2}}cos(\alpha)+\frac{y}{\sqrt{x^2+y^2}}sin(\alpha))=a\\ xcos(\alpha)+ysin(\alpha)=a\\$