# Find an equation for the line tangent to the curve at the point defined by the given value of t ...

## Question:

Find an equation for the line tangent to the curve at the point defined by the given value of {eq}t{/eq}

• {eq}x=\csc t, y=6 \cot t, t =\frac{\pi}{3}{/eq}

A) {eq}y=-12x+6\sqrt3{/eq} {eq}\enspace \enspace \enspace {/eq}B) {eq}y=2\sqrt3x-12{/eq} {eq}\enspace \enspace \enspace {/eq}C) {eq}y=12x-6\sqrt3{/eq} {eq}\enspace \enspace \enspace {/eq}D) {eq}y=12x+2\sqrt3{/eq}

(Choose the correct one)

## Parametrics and Cartesian Equations:

To move from an equation in parametrics to an equation in Cartesian we have to eliminate the parameter and then rewrite it in terms of the Cartesian variables.

First, with the value of the function and the first derivative, we obtain the tangent line in parametrics:

{eq}x = \csc t,y = 6\cot t,t = \frac{\pi }{3}\\ t = \frac{\pi }{3} \to x = \csc \frac{\pi }{3} = \frac{2}{{\sqrt 3 }},y = 6\cot \frac{\pi }{3} = 6 \cdot \frac{1}{{\sqrt 3 }} = \frac{6}{{\sqrt 3 }}\\ \\ x' = - \frac{1}{{{{\sin }^2}t}}\cos t,y' = 6\left( { - \frac{1}{{{{\sin }^2}t}}} \right)\\ t = \frac{\pi }{3} \to x' = - \frac{1}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}\left( {\frac{1}{2}} \right) = - \frac{2}{3},y' = 6\left( { - \frac{1}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} \right) = - 8\\ \\ \left\{ \begin{array}{l} x = \frac{2}{{\sqrt 3 }} - \frac{2}{3}t\quad \left( 1 \right)\\ y = \frac{6}{{\sqrt 3 }} - 8t\quad \left( 2 \right) \end{array} \right. {/eq}

Once the parameter is eliminated, we obtain the equation of the tangent line:

{eq}\frac{1}{{12}}\left( 2 \right) - \left( 1 \right)\\ \frac{1}{{12}}y - x = \frac{1}{{12}}\left( {\frac{6}{{\sqrt 3 }} - 8t} \right) - \left( {\frac{2}{{\sqrt 3 }} - \frac{2}{3}t} \right)\\ \frac{1}{{12}}y - x = \frac{{ - 3}}{{2\sqrt 3 }}\\ \left[ {\frac{1}{{12}}y - x = \frac{{ - 3}}{{2\sqrt 3 }}} \right]12\\ y - 12x = - \frac{{18}}{{\sqrt 3 }}\\ y = 12x - \frac{{18}}{{\sqrt 3 }}\\ y = 12x - 6\sqrt 3 {/eq}

So, the correct option is C) 