# Find the equation in x and y for the line tangent to the curve in polar coordinates r = 2 \cos...

## Question:

Find the equation in {eq}\,x\, {/eq} and {eq}\,y\, {/eq} for the line tangent to the curve in polar coordinates {eq}\,r = 2\cos 2\theta\, {/eq} at the value {eq}\displaystyle\,\theta = \frac{\pi}{2} {/eq}.

## Tangent to a Polar Curve:

The slope of the tangent to the polar curve with equation

{eq}r = r(\theta) {/eq}

is calculated according to the following formula

{eq}\displaystyle \frac{dy}{dx} =\frac{ \frac{dr}{d\theta} \sin \theta + r \cos \theta } { \frac{dr}{d\theta} \cos \theta - r \sin \theta }. {/eq}

## Answer and Explanation:

Given the curve in polar coordinates

{eq}\,r = 2\cos 2\theta\, {/eq}

the slope of the tangent at the value {eq}\displaystyle\,\theta = \frac{\pi}{2} {/eq} is calculated as follows

{eq}\displaystyle m=\frac{dy}{dx} =\frac{ \frac{dr}{d\theta} \sin \theta + r \cos \theta } { \frac{dr}{d\theta} \cos \theta - r \sin \theta } \\ \displaystyle =\frac{- 4\sin 2\theta \sin \theta + 2\cos 2\theta \cos \theta } { -4\sin 2\theta \cos \theta - 2 \cos2\theta \sin \theta } \\ \displaystyle =\frac{+ 2\cos 2\theta \cos \theta } { -4\sin 2\theta \cos \theta - 2 \cos2\theta \sin \theta } \\ \Rightarrow m(\frac{\pi}{2}) = 0. {/eq}

The angle {eq}\theta=\frac{\pi}{2} {/eq} corresponds to the point with cartesian coordinates

{eq}x=r\cos\theta = 0 \\ y=r\sin\theta = 2\cos 2\theta\, \sin\theta = -2. {/eq}

Then the equation in {eq}\,x\, {/eq} and {eq}\,y\, {/eq} for the line tangent to the curve a the point (0,-2) is

{eq}y = -2. {/eq}

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