How do you convert $6 = {(5 x + 4 y)}^{2} - 7 x$ $6=(5x+4y)^2-7x$ into polar form?

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Answer 1 · 2018-07-16T04:44:10

$r^{2} {(4 cos θ + 5 sin θ)}^{2} - 7 r cos θ - 6 = 0$ $r^2(4\cos \theta + 5\sin \theta)^2-7r\cos \theta-6=0$ .

Just remember these coordinate conversions:

$x = r cos θ$ $x=r\cos \theta$

$y = r sin θ$ $y=r\sin \theta$

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

So the given equation is converted to

$6 = {(4 r cos θ + 5 r sin θ)}^{2} - 7 r cos θ$ $6=(4r\cos \theta + 5r\sin \theta)^2-7r\cos \theta$ .

We can rearrange this to:

$r^{2} {(4 cos θ + 5 sin θ)}^{2} - 7 r cos θ - 6 = 0$ $r^2(4\cos \theta + 5\sin \theta)^2-7r\cos \theta-6=0$ .

How do you convert 6=(5x+4y)^2-7x6=(5x+4y)2−7x6=(5x+4y)^2-7x into polar form?