Question 635288
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.25x^2\ +\ y^2\ =\ 9]


Find *[tex \LARGE (x,y)] such that *[tex \LARGE \frac{dy}{dx}\ =\ 1]


Differentiate implicitly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.5x\ +\ 2y\frac{dy}{dx}\ =\ 0]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \frac{-0.5x}{2y}]


But if *[tex \LARGE \frac{dy}{dx}\ =\ 1], then *[tex \LARGE \frac{-0.5x}{2y}\ =\ 1], hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{x}{4}]


Substituting into the original equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2}{4}\ +\ \left(\frac{x}{4}\right)^2\ =\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x^2}{4}\ +\ \frac{x^2}{16}\ =\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x^2\ =\ 144]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\frac{12\sqrt{5}}{5}]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{x}{4}]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \mp\frac{3\sqrt{5}}{5}]


And the two points are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-\frac{12\sqrt{5}}{5},\,\frac{3\sqrt{5}}{5}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{12\sqrt{5}}{5},\,-\frac{3\sqrt{5}}{5}\right)]


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