How do you convert $3 = {(7 x - 5 y)}^{2} - y$ $3=(7x-5y)^2-y$ into polar form?

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How do you convert $3 = {(7 x - 5 y)}^{2} - y$ $3=(7x-5y)^2-y$ into polar form?

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Answer 1 · 2018-04-23T09:37:58

Put $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$

Explanation:

$3 = {(7 x - 5 y)}^{2} - y$ $3=(7x-5y)^2 - y$ ...............(Given equation)

Put $x = r cos θ$ $x=rcostheta$ and $y = r sin θ$ $y=rsintheta$ ; we get :-

$\Rightarrow 3 = {(7 r cos θ - 5 r sin θ)}^{2} - r sin θ$ $rArr3=(7rcostheta-5rsintheta)^2-rsintheta$

$\Rightarrow 3 = 49 r^{2} {cos}^{2} θ + 25 r^{2} {sin}^{2} θ - 70 r^{2} cos θ . sin θ - r sin θ$ $rArr3=49r^2cos^2theta+25r^2sin^2theta-70r^2costheta.sintheta-rsintheta$

$\Rightarrow 3 = 24 r^{2} {cos}^{2} θ + (25 r^{2} {cos}^{2} θ + 25 r^{2} {sin}^{2} θ) - 70 r^{2} cos θ . sin θ - r sin θ$ $rArr3=24r^2cos^2theta+(25r^2cos^2theta+25r^2sin^2theta)-70r^2costheta.sintheta-rsintheta$

$\Rightarrow 3 = 24 r^{2} {cos}^{2} θ + 25 r^{2} - 70 r^{2} cos θ . sin θ - r sin θ$ $rArr3=24r^2cos^2theta+25r^2-70r^2costheta.sintheta-rsintheta$

$:.25r^2+24r^2cos^2theta-70r^2costheta.sintheta-rsintheta-3=0$

is the Polar form

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How do you convert $3 = {(7 x - 5 y)}^{2} - y$ $3=(7x-5y)^2-y$ into polar form?

1 Answer

Explanation:

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How do you convert $3 = {(7 x - 5 y)}^{2} - y$ $3=(7x-5y)^2-y$ into polar form?