Question 282460
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For the general parabola with axis parallel (or coincident with) the *[tex \Large y]-axis, namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ ax^2\ +\ bx\ +\ c]


the directrix is the line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ c\ -\ \frac{b^2\,+\,1}{4a}]


the focus is the point:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{-b}{2a},\,c\ -\ \frac{b^2\,-\,1}{4a}\right)]


and the roots are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}]


For your problem, *[tex \Large a\ =\ 1], *[tex \Large b\ =\ -5], and *[tex \Large c\ =\ 4]


Just plug in the coefficient values and do the arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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