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Two tangents to the parabola $y^2= 8x$ meet the tangent at its vertex in the points $P$ and $Q$. If $|PQ| = 4$, prove that the locus of the point of the intersection of the two tangents is $y^2 = 8 (x + 2)$. I take the point of intersection $(x,y)$ . But after that how can I write equation of the tangents?

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  • $\begingroup$ I doubt this should be tagged a paradox. Also could you elaborate some more on what you have tried so far? What are your thoughts? Adding details about that will improve the exchange here. $\endgroup$ – String Dec 15 '15 at 12:41
  • $\begingroup$ Let the eq of first tangent be y=mx +a/m . Hence it intersect the tangent at its vertex i.e T(0,a/m) so T` is (0,4-a/m) and we got the equation of both lines as let the intersection point be (x,y) but hot to eliminate m $\endgroup$ – Koolman Dec 15 '15 at 12:50
  • $\begingroup$ It's not clear to me what you mean by $T(0,a/m)$. I'm afraid there may be a bit of a language barrier here as well. What do you mean by the two tangents "meet[ing] the tangent at its vertex"? Do you mean to consider the tangents at two points $P$ and $Q$ along the parabola $y = x^2$? $\endgroup$ – jdc Dec 15 '15 at 19:45
  • $\begingroup$ The slope of the tangent line at (x,y) is 4/y . $\endgroup$ – zyx Dec 16 '15 at 17:49
  • $\begingroup$ Yes, parabola is a conic section, but this problem is not about a conic section itself, nor the parabola section. I'd suggest changing the title to better reflect the problem, e.g. 'Parabola tangents intersection locus' or 'Where do tangents to parabola intersect?' or something like that... $\endgroup$ – CiaPan Mar 21 '17 at 7:26
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Use the fact that perpendicular from the focus on any tangent to a parabola meets the tangent at the vertex. So let$ty=x+at^2$ be the equation of tangent. Equation of line perpendicular from focus is $y=-tx+ta$ Solve it with the equation of tangent. Get the points. Use the fact that the distance is given and the point of intersection of tangents is $(at_1t_2, a(t_1+t_2)$

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This is to separate the geometric statement implied in the problem from the particular numbers used.

The question claims that if we slide a line segment of constant length $L$ along a parabola, the locus of the intersection point of the tangents at the ends of the segment is another parabola. In addition, the second parabola is a translation of the first along the axis of symmetry. The magnitude of the translation is apparently some simple function of $L$ that sends $2$ to $4$.

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  • $\begingroup$ I think you mean: "... slide a line segment of constant length $L$ along a tangent to a parabola ...", where "a tangent" is specifically the tangent at the vertex (although I think this might work for any tangent line). $\endgroup$ – Blue Dec 16 '15 at 18:07
  • $\begingroup$ Ah. I read it as jdc did in the comments under the question, but your version seems like a more faithful translation. $\endgroup$ – zyx Dec 16 '15 at 18:34

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