Actually the definition of a parabola is: a set of points is the plane that are equally distance from a fix point(the focus) and a fix line (the directrix).The graph of a quadratic function is called a parabola. The x^2 term leads to this definition. Is this the only reason why it's called a parabola? Is there a more mathematical reason? How important is the study of the parabola in calculus?
A definition of the parabola that I like is the set of all points equidistant from a line (directrix) and a point not on the line (focus). If we're talking about a function, then the directrix is obviously horizontal.
Let the directrix be the line \(y=k\) and the focus be the point \((x_f,y_f)\). Using the definition above, we may then state:
[MATH]\left(x-x_f\right)^2+\left(y-y_f\right)^2=(y-k)^2[/MATH]
[MATH]\left(x-x_f\right)^2+y^2-2y_fy+y_f^2=y^2-2ky+k^2[/MATH]
[MATH]\left(x-x_f\right)^2+y_f^2-k^2=2y\left(y_f-k\right)[/MATH]
[MATH]y=\frac{\left(x-x_f\right)^2+y_f^2-k^2}{2\left(y_f-k\right)}[/MATH]
Now, we know the axis of symmetry will be the line through the focus and perpendicular to the directrix, which is \(x=x_f\) and the vertex will be at the point on the axis of symmetry midway between the focus and directrix which is \((h,k)=\left(x_f,\dfrac{k+y_f}{2}\right)\), and so we obtain:
[MATH]y=\frac{1}{2\left(y_f-k\right)}\left(x-h\right)^2+k[/MATH]
We then see that if the focus is above the directrix, the parabola opens upwards, and if the focus is below the directrix, the parabola opens downwards.
Actually the definition of a parabola is: a set of points is the plane that are equally distance from a fix point(the focus) and a fix line (the directrix).
In standard form the directrix is horizontal. So if the focus is above the line the parabola must open upwards, and the opposite if the focus is below the directrix.
I like it, but you probably didn't intend to use symbol k for both the directrix y-intercept and the vertex y-coordinate.... Let the directrix be the line \(y=k\) ...
[MATH]y=\frac{\left(x-x_f\right)^2+y_f^2-k^2}{2\left(y_f-k\right)}[/MATH]\((h,k)=\left(x_f,\dfrac{k+y_f}{2}\right)\)
I like it, but you probably didn't intend to use symbol k for both the directrix y-intercept and the vertex y-coordinate.
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