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If a triangle is formed by any three tangents of the parabola $y^2=4ax$, two of whose vertices lie on the parabola $x^2=4by$, then find the locus of the third vertex.

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    $\begingroup$ If I remember correctly, I had solved this a few months ago. and the answer (surprisingly) turned out to be the same parabola. It was done by some manipulation of quadratics in the parameter t. If I get some free time tomorrow, I'll try it again. $\endgroup$
    – Sawarnik
    Nov 27 '16 at 20:53
  • $\begingroup$ May be this one $\endgroup$ Nov 27 '16 at 22:39
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You can easily show that tangents to $y^2=4ax$ at points $t_1$ and $t_2$ intersect at the point $(at_1 t_2, a(t_1 + t_2))$. Now we are given three points $t_1, t_2, t_3$ satisfying the two relations:

$$(at_1 t_2)^2=4ab(t_1 + t_2)$$ $$(at_3 t_2)^2=4ab(t_3 + t_2)$$

And we need to find the locus of the intersection of other two tangents, that is, a relation between $x=at_1 t_3$ and $y=a(t_1 + t_3)$.

Note that, by rearranging the above equations, both $t_1$ and $t_3$ satisfy the quadratic equation $t^2(at_2^2) - 4bt - 4bt_2=0$ (assuming $t_2$ to be constant). Thus we can find $t_1 t_3$ and $t_1 + t_3$ directly from this equation which are the sum and product of roots of the quadratic. Thus:

$$t_1 + t_3 = \frac{4b}{at_2^2}$$ $$t_1t_3 = \frac{4b}{at_2}$$

Now you can substitute $x=at_1 t_3$ and $y=a(t_1 + t_3)$ in the above relations, and straightforward manipulation leaves you with the locus $x^2=4by$.

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  • $\begingroup$ The fact that you use $x$ to denote $at_1t_3$ and then use $x$ to denote $t_1$ resp. $t_3$ is somewhat confusing. Perhaps you can write the latter as $t_i$ or $\tau$ or something like that? $\endgroup$
    – MvG
    Nov 28 '16 at 15:57
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    $\begingroup$ @MvG Yes, those were completely unrelated. Its just out of habit that we take the variable of a quadratic to be $x$. I didn't see that it would conflict with the $x$ coordinate here. Corrected. $\endgroup$
    – Sawarnik
    Nov 29 '16 at 2:55
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Here is one approach. Use $t$ as a parameter. Find the tangent to the first parabola in $(at^2,2at)$ or some such. Intersect that line with the second parabola, the two points of intersection being the two vertices you described. Compute the other tangents (i.e. different from the first tangent) from these to the first parabola, and intersect them. Eliminate $t$ from the resulting coordinates to obtain an implicit description.

Doing the computation here in Sage I can confirm what Sawarnik remembered: The locus will be the parabola $x^2=4by$. My parametrization above leads to coordinates $(-4b/t, 4b/t^2)$ for the third vertex. Along the way I had some nasty square roots, but there has to be a way to avoid these.

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