Can we find the locus of a midpoint created by a parabola?
Question
Show that the tangent to the curve at the point where is
This tangent meets the -axis at and the -axis at . The mid-point of is .
Find the co-ordinates of in terms of and hence deduce the equation of the locus of as varies.
Suggestion
To find the gradient, it might help to expand the bracket.
Given two points in the plane whose coordinates are and , can we write down the coordinates of their midpoint?
Solution
We have , and so . (If you know the chain rule, you can alternatively use this here.)
This tells us that when , and . So the gradient of the tangent is .
We can write the equation of the tangent as , and we know that the point is on the line, so
Thus the equation is
or, after rearranging,
as required.
From ((given-form-for-tangent)), the coordinates of are just . Similarly, the -coordinate of is given by and so . Thus, the coordinates of are .
The midpoint of the line segment is thus
To find the locus of , we need to eliminate here. So if is on the locus of ,
Now we have the locus of as varies; a vertex-up parabola, with its maximum at the origin. You can observe the locus in the following applet by varying .
[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/dDg5zVQh/width/513/height/620/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.
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