Agaiepsh

2021-11-15

Consider the following three equations:

$A)r=\frac{6}{3+2\mathrm{cos}\left(\theta \right)}$

Troy Lesure

Beginner2021-11-16Added 26 answers

Step 1

Solution:

$A)\text{}r=\frac{6}{3+2\mathrm{cos}\theta}$

$\Rightarrow r(3+2\mathrm{cos}\theta )=6$

$3r+2r\mathrm{cos}\theta =6$

$3\sqrt{{x}^{2}+{y}^{2}}+2x=6$

$\Rightarrow 3\sqrt{{x}^{2}+{y}^{2}}=6-2x$

Squaring both the sides

$9({x}^{2}+{y}^{2})=36-24x+4{x}^{2}$

$9{x}^{2}-4{x}^{2}-24x+9{y}^{2}=36$

$5{x}^{2}-24x+9{y}^{2}=36$

$5({x}^{2}-\frac{24}{5}x)+9{y}^{2}=36$

$5({x}^{2}-2\times \frac{12}{5}x+{\left(\frac{12}{5}\right)}^{2})+9{y}^{2}=36+5{\left(\frac{12}{5}\right)}^{2}$

$5{(x-\frac{12}{5})}^{2}+9{y}^{2}=36+\frac{144}{5}$

$5{(x-\frac{12}{5})}^{2}+9{y}^{2}=\frac{324}{5}$

Dividing throughout by$\frac{324}{5}$

$\frac{5{(x-\frac{12}{5})}^{2}}{\frac{324}{5}}+\frac{9{y}^{2}}{\frac{324}{5}}=1$

$\Rightarrow \frac{{(x-\frac{12}{5})}^{2}}{\frac{324}{25}}+\frac{{y}^{2}}{\frac{324}{45}}=1$

$\frac{{(x-\frac{12}{5})}^{2}}{{\left(\frac{18}{5}\right)}^{2}}+\frac{{y}^{2}}{{\left(\frac{18}{\sqrt{45}}\right)}^{2}}=1$

Hence, this is the equation of an ellipse of the form

Solution:

Squaring both the sides

Dividing throughout by

Hence, this is the equation of an ellipse of the form

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