How do you convert $2 = {(3 x + 7 y)}^{2} - x$ $2=(3x+7y)^2-x$ into polar form?

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How do you convert $2 = {(3 x + 7 y)}^{2} - x$ $2=(3x+7y)^2-x$ into polar form?

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Answer 1 · 2017-12-17T17:28:46

$29 r^{2} - 20 r^{2} cos (2 θ) + 21 r^{2} sin (2 θ) - r cos (θ) - 2 = 0$ $29r^2-20r^2cos(2theta)+21r^2sin(2theta)-rcos(theta)-2=0$

Explanation:

$2 = {(3 x + 7 y)}^{2} - x$ $2=(3x+7y)^2-x$

$9 x^{2} + 42 x y + 49 y^{2} - x = 2$ $9x^2+42xy+49y^2-x=2$

After using $x = r cos (θ)$ $x=rcos(theta)$ and $y = r sin (θ)$ $y=rsin(theta)$ transormation,

$9 r^{2} {(cos (θ))}^{2} + 42 r^{2} \cdot cos (θ) \cdot sin (θ) + 49 r^{2} {(sin (θ))}^{2} - r cos (θ) = 2$ $9r^2(cos(theta))^2+42r^2*cos(theta)*sin(theta)+49r^2(sin(theta))^2-rcos(theta)=2$

$9 r^{2} \cdot \frac{1 + cos (2 θ)}{2} + 21 r^{2} \cdot sin (2 θ) + 49 r^{2} \frac{1 - cos (2 θ)}{2} - r cos (θ) = 2$ $9r^2*(1+cos(2theta))/2+21r^2*sin(2theta)+49r^2(1-cos(2theta))/2-rcos(theta)=2$

$29 r^{2} - 20 r^{2} cos (2 θ) + 21 r^{2} sin (2 θ) - r cos (θ) - 2 = 0$ $29r^2-20r^2cos(2theta)+21r^2sin(2theta)-rcos(theta)-2=0$

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How do you convert $2 = {(3 x + 7 y)}^{2} - x$ $2=(3x+7y)^2-x$ into polar form?

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How do you convert 2=(3x+7y)^2-x2=(3x+7y)2−x2=(3x+7y)^2-x into polar form?

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How do you convert $2 = {(3 x + 7 y)}^{2} - x$ $2=(3x+7y)^2-x$ into polar form?