Question 308473
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If there is a squared x term and not a squared y term (or vice versa) then you have a parabola.  If x is the squared term and there is no xy term, then the axis is a vertical line.  If y is the squared term and there is no xy term, then the axis is a horizontal line.  If there is an xy term, then the axis is an oblique line.


If you have both an x squared term and a y squared term and the coefficients on these two terms are equal, then you have a circle.


If the coefficients are unequal, but the signs are the same, then you have an ellipse.  The term with the smaller coefficient (or larger denominator on the coefficient) identifies the major axis.  Unless there is an xy term in which case the axes of the ellipse are oblique to the coordinate axes.


If the coefficients are unequal and the signs are different, then you have a hyperbola.


So, all three of your "I thinks" are correct.


For your last one, plot your parabola on a graph.  Plot a point at (-2,0), another at (2,0), and a third one at (0,-1.5).


Now we need the equation of a parabola that passes through all three points.


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


Is the general equation of a parabola.  But we know that the parabola we want must pass through (-2,0), so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(-2)^2\ +\ b(-2)\ +\ c\ =\ 0]


and it must pass through (2, 0), so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(2)^2\ +\ b(2)\ +\ c\ =\ 0]


and it must pass through (0, -1.5), so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(0)^2\ +\ b(0)\ +\ c\ =\ -1.5]


Which we can rewrite as the system of equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ +\ c\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ +\ c\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0a\ +\ 0b\ +\ c\ =\ -1.5]


Ah, ha...*[tex \Large c\ =\ -1.5]


So now we have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ -\ 2b\ =\ 1.5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4a\ +\ 2b\ =\ 1.5]


Add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8a\ +\ 0b\ =\ 3]


And we have *[tex \Large a\ =\ \frac{3}{8}]


I'll leave it to you to verify that *[tex \Large b\ =\ 0]


So, let's write our equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{3}{8}x^2\ -\ \frac{3}{2}]


Next, let's put it into *[tex \Large (x\ -\ h)^2\ =\ 4p(y\ -\ k)] form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ \frac{8}{3}(y\ +\ \frac{3}{2})]


So,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4p\ =\ \frac{8}{3}]


Which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p\ =\ \frac{2}{3}]


And that means our focus is *[tex \Large \frac{2}{3}] ft distant from the vertex  which we established at (0,-1.5).


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{3}{2}\ +\ \frac{2}{3}\ =\ -\frac{5}{6}]


And the focus is at *[tex \Large (0,-\frac{5}{6})], which is 8 inches (2/3 foot) along the axis from the lowest point of the dish reflector.


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