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You are given two points (2,3) and (3,1) on Parabola from where 2 tangents are drawn intersecting at (-1,-2) . Find the locus of Parabola

This question is supposed to have some trick or some clever approach I have tried really hard but I am failing to see any gimmicky solution

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    Do you know something about the orientation of the parabola? Is the axis vertical? – Emilio Novati Jan 28 '20 at 15:59
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    Nothing else is given I believe – total dependent random choice Jan 28 '20 at 16:44
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    And nothing else is needed either. You can put the general equation and substitute the points there, take derivatives and equate to the slopes of the tangent lines to find all the coefficients. Let me try another way, though. The point $(3,1)+((2,3)-(-1,-2))=(6,6)$ and the point $(-1,-2)$ determine a line that is parallel to the axis of the parabola. Drawing a line passing through $(2,3)$ and parallel to the axis gives you a line that if reflected with respect to the perpenticular to the tangent at $(2,3)$ gives a line passing though the focus. Doing this for $(3,1)$ gives another line ... – OscarRascal Jan 28 '20 at 17:08
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    ... passing through the focus. So, you get the focus. With the focus and the axis you have the parabola. – OscarRascal Jan 28 '20 at 17:09
  • @OscarRascal I’m sorry but exactly what did you to do arrive at the point (6,6)? – total dependent random choice Jan 28 '20 at 17:36
  • Apply the addition and subtraction component-wise. In other words, fin the fourth point that forms a parallelogram. See here. – OscarRascal Jan 28 '20 at 17:47
  • It does. Thank you @OscarRascal Really appreciate your help – total dependent random choice Jan 28 '20 at 17:56
  • @OscarRascal Focus and axis direction by themselves aren’t enough to determine the parabola, but combined with a known point and a bit more work, they are. – amd Jan 28 '20 at 19:16
  • @amd Do you see "alone", "themselves", "only"? Then scram, don't waste my time telling me what I already know. – OscarRascal Jan 28 '20 at 19:25

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