Agaiepsh

2021-11-15

Consider the following three equations:
$A\right)r=\frac{6}{3+2\mathrm{cos}\left(\theta \right)}$

Troy Lesure

Step 1
Solution:

$⇒r\left(3+2\mathrm{cos}\theta \right)=6$
$3r+2r\mathrm{cos}\theta =6$
$3\sqrt{{x}^{2}+{y}^{2}}+2x=6$
$⇒3\sqrt{{x}^{2}+{y}^{2}}=6-2x$
Squaring both the sides
$9\left({x}^{2}+{y}^{2}\right)=36-24x+4{x}^{2}$
$9{x}^{2}-4{x}^{2}-24x+9{y}^{2}=36$
$5{x}^{2}-24x+9{y}^{2}=36$
$5\left({x}^{2}-\frac{24}{5}x\right)+9{y}^{2}=36$
$5\left({x}^{2}-2×\frac{12}{5}x+{\left(\frac{12}{5}\right)}^{2}\right)+9{y}^{2}=36+5{\left(\frac{12}{5}\right)}^{2}$
$5{\left(x-\frac{12}{5}\right)}^{2}+9{y}^{2}=36+\frac{144}{5}$
$5{\left(x-\frac{12}{5}\right)}^{2}+9{y}^{2}=\frac{324}{5}$
Dividing throughout by $\frac{324}{5}$
$\frac{5{\left(x-\frac{12}{5}\right)}^{2}}{\frac{324}{5}}+\frac{9{y}^{2}}{\frac{324}{5}}=1$
$⇒\frac{{\left(x-\frac{12}{5}\right)}^{2}}{\frac{324}{25}}+\frac{{y}^{2}}{\frac{324}{45}}=1$
$\frac{{\left(x-\frac{12}{5}\right)}^{2}}{{\left(\frac{18}{5}\right)}^{2}}+\frac{{y}^{2}}{{\left(\frac{18}{\sqrt{45}}\right)}^{2}}=1$
Hence, this is the equation of an ellipse of the form

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