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If the locus of circumcentre of a variable triangle having sides $x=3$, the X axis, and $px+qy=4$, where (p,q) lies on the parabola $x^2=4y$ is the given curve, then find the axis about which the curve is symmetric.

This is obviously a very long and tedious question, but I have solved it right (hopefully) and arrived at the equation $$(x-\frac 94)^2=\frac 12 (y-\frac 98)$$

My question is, have I done this right? And also I am not able to find the axis of symmetry, so please help me with that. Thanks

Aditya
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1 Answers1

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Note that $p^2=4q$ and the third side is $px+\frac{p^2}4y=4$. Then, the intersection of the $y=0$ and the third side is the vertex $(\frac4p,0)$ and the intersection of $x=3$ and the third side is the vertex $(3, \frac{16}{p^2}-\frac{12}p)$.

Since the two sides $y=0$ and $x=3$ are perpendicular to each other, the circumcenter is the midpoint of the two vertexes found above, i.e

$$x = \frac2p+\frac32,\>\>\>\>\>y=\frac{8}{p^2}-\frac{6}p$$

Eliminate $p$ to obtain the equation of the locus

$$ 2\left(x-\frac94\right)^2=y + \frac98$$

which differs from your result by a sign. The above locus is an upright parabola with its vertex at $(\frac94,-\frac98)$. Thus, the axis of symmetry is $x= \frac94$.

Quanto
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