MX1 Locus (1 Viewer)

spatula232 · Aug 6, 2015

Was asked to find the locus of the midpoint, found it to be x^2=2a(y-4a)
The answers said it was a parabola with Vertex at (0,4a)
Doesn't it have to be in the form x^2=4ay? Or would I just leave it as is?

Thanks,
Spatula

InteGrand · Aug 6, 2015

spatula232 said:
Was asked to find the locus of the midpoint, found it to be x^2=2a(y-4a)
The answers said it was a parabola with Vertex at (0,4a)
Doesn't it have to be in the form x^2=4ay? Or will it work out to be the same?

Thanks,
Spatula

$The equation of the ``vertical'' parabola (by which I mean that its axis of symmetry is parallel to the $y$-axis) is of the form $x^2 = 4Ay$ if and only if the vertex is at the origin ($A$ being a non-zero constant). In your case, you have $x^2 = 2a(y-4a)$, so the vertex is $(0,4a)$. (Recall that the general form of the equation for a vertical parabola is $(x-h)^2 = 4a(y-k)$, where $(h,k)$ will be the vertex and $|a|$ is the focal length.)$

spatula232 · Aug 6, 2015

x^2 = 2a(y-4a)
Should that be 4a or 2a [and therefore change make it x^2=4a(y/2 - 2a)], or does it not matter?
Apologies, I wasn't very clear

InteGrand · Aug 6, 2015

spatula232 said:
x^2 = 2a(y-4a)
Should that be 4a or 2a, or does it not matter?
Apologies, I wasn't very clear

$Doesn't matter, since $2a$ is of the form $4A$, with $A = \frac{2a}{4}$. So the focal length of that parabola is \left |\frac{2a}{4}\right | = \left \frac{|a|}{2}.$$

InteGrand · Aug 6, 2015

spatula232 · Aug 6, 2015

Awesome thanks

MX1 Locus (1 Viewer)

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