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Tangents drawn to the parabola y2=4ax at the points P and Q intersect at T. If triangle TPQ is equilateral, then find the side length of this triangle.
APPROACH
P (at12 ,2at1) ; Q(at22 ,2at2) ; T (at1t2,a(t1 + t2))
I then applied the distance formula and equated two sides to get a relation, but in the end I couldn't get the answer.

Ava
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2 Answers2

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Using the fact that $$\text{$\triangle{\alpha\beta\gamma}$ is an equilateral triangle $\iff \alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha=0$}$$ where $\alpha,\beta,\gamma$ are complex numbers, we have $$(at_1^2+2at_1i)^2+(at_2^2+2at_2i)^2+(at_1t_2+a(t_1+t_2)i)^2-(at_1^2+2at_1i)(at_2^2+2at_2i)-(at_2^2+2at_2i)(at_1t_2+a(t_1+t_2)i)-(at_1t_2+a(t_1+t_2)i)(at_1^2+2at_1i)=0,$$ i.e. $$a^2t_1^4-4a^2t_1^2+a^2t_2^4-4a^2t_2^2+a^2t_1^2t_2^2-a^2(t_1+t_2)^2-a^2t_1^2t_2^2+4a^2t_1t_2-a^2t_1t_2^3+2a^2t_2(t_1+t_2)-a^2t_1^3t_2+2a^2t_1(t_1+t_2)+(4a^2t_1^3+4a^2t_2^3+2a^2t_1t_2(t_1+t_2)-2a^2t_1^2t_2-2a^2t_1t_2^2-a^2t_2^2(t_1+t_2)-2a^2t_1t_2^2-2a^2t_1^2t_2-a^2t_1^2(t_1+t_2))i=0,$$ i.e. $$a^2(t_1-t_2)^2(t_1^2+t_1t_2+t_2^2-3)+3a^2(t_1-t_2)^2(t_1+t_2)i=0$$ So, we have $$t_1^2+t_1t_2+t_2^2-3=t_1+t_2=0\quad\Rightarrow\quad t_1t_2=-3$$ Hence, the side length of the equilateral triangle is $$\sqrt{(at_1^2-at_2^2)^2+(2at_1-2at_2)^2}=\sqrt{4a^2((t_1+t_2)^2-4t_1t_2)}=\color{red}{4\sqrt{3}\ |a|}.$$

mathlove
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  • Why can't we use distance formula ? – Ava Mar 28 '16 at 12:58
  • @Ava: I think we can. Having $PQ^2=PT^2,PQ^2=QT^2$ and factoring, you'll get $t_1+t_2=0,t_1t_2=-3$ as I got. But that would be very tedious (I admit that the way in my answer is also tedious.) Do you want an answer using distance formula? – mathlove Mar 28 '16 at 13:10
  • what do you mean by factoring? Does it mean making the LHS and RHS into factors? – Ava Mar 28 '16 at 13:52
  • Thank you, I have solved it. – Ava Mar 28 '16 at 14:14
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Since the question states, that there is a unique answer (else you could not use symmetry so easily), there is a much easier approach than that you tried.

Look at the angles, they have to be 60°, it's more handy than the sides. Now you can compute the derivative $\frac{\mathrm dx}{\mathrm dy}$ for any y and look where it gives an angle of 60° (i.e. the derivative is $\sqrt3$)

... But your approach should also work, if you use the symmetry argument (there has to be an equilateral triangle with $T$ on the x-axis, since if you move this point away from (0, 0) the angle near $T$ continuously decreases and has to pass 60° -- so if there's assumed a unique solution, we can take this) : so $t_1=t_2$

Ilja
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  • I am not using symmetry. I am using the basic concenpt of parabola that if there are two tangents at t1 and t2, they would meet at (at1t2, a(t1+t2)). And if t1=t2, the side would become 0. This is not the ans. – Ava Mar 28 '16 at 12:45
  • there is something wrong here... this formula should be valid if the $t_[1,2]$ are x-coordinates in a normal parabola, but your example is rotated differently, in y-direction – Ilja Mar 28 '16 at 12:56
  • The parabola is not rotated. The parametric coordinates of standard parabola are (at^2,2at) – Ava Mar 28 '16 at 13:00
  • well, this is strange, then we know different standards :) I would say, that $y=x^2$ is the normal form, and not $x=y^2$ – Ilja Mar 28 '16 at 13:02