Question 625669
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Differentiate the equation implicitly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\left(7x^2\ -\ 6\sqrt{3}xy\ +\ 13y^2\ -\ 16\right)\ =\ \frac{d0}{dx}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d\left(7x^2\right)}{dx}\ -\ \frac{d\left(6\sqrt{3}xy\right)}{dx}\ +\ \frac{d\left(13y^2\right)}{dx}\ -\ \frac{d(16)}{dx}\ =\ 0]


Use the Chain Rule and the Product Rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 14x\ -\ 6\sqrt{3}\left(x\frac{dy}{dx}\ +\ y\frac{dx}{dx}\right)\ +\ 13y\frac{dy}{dx}\ -\ 0\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 14x\ -\ 6\sqrt{3}x\frac{dy}{dx}\ -\ 6\sqrt{3}y\ +\ 13y\frac{dy}{dx}\ =\ 0]


Now take out a common factor of 2 and then solve for *[tex \LARGE \frac{dy}{dx}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \frac{3\sqrt{3}y\ -\ 7x}{13y\ -\ 3\sqrt{3}x}]


Do not succumb to the temptation to rationalize your denominator at this point.  We have an expression for the tangent to your ellipse at any point,  all that is needed now is to evaluate this expression at the given point.  Substitute the coordinates of the given point for *[tex \LARGE x] and *[tex \LARGE y] in the expression for *[tex \LARGE \frac{dy}{dx}] and do the arithmetic.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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