anonymous
  • anonymous
Help and assurance, once more :) P is a point on the parabola given parametrically by x=at^2, y=2at, where a is a constant. Let S be the point (a,0), Q be the point (-a, 2at) and T be the point where the tangent at P to the parabola crosses the axis of symmetry of the parabola. (a) Show that SP=PQ=QT=ST=at^2+a
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
x'=2at y'=2a dy/dx= 1/t
anonymous
  • anonymous
equation of line = y= 1/tx+b 2at=(1/t)(at^2) +b 2at=(at) +b b=at y=(1/t)x +at
anonymous
  • anonymous
Now what?

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anonymous
  • anonymous
Find the intersection of axis of symmetry and Tangent
anonymous
  • anonymous
How do I do that?
anonymous
  • anonymous
Axis of symmetry is nonetheless the x-axis or y=0. y = (1/t)x + at, y =0 x = -at^2 => point of intersection is (-at^2,0)
anonymous
  • anonymous
Now use the distance formula to get the proof.
anonymous
  • anonymous
Ok, thank you. There's a second part to the question, which I'll post up. Can you just show me the proof? I know the distance formula... but just need some assurance
anonymous
  • anonymous
@Ishaan94
apoorvk
  • apoorvk
help still needed??.... proof for distance formula?
anonymous
  • anonymous
Yes please
anonymous
  • anonymous
"... S be the point (a,0), Q be the point (-a, 2at) and T ... " "P is a point on the parabola given parametrically by x=at^2, y=2at ..." "... Show that SP=PQ=QT=ST=at^2+a" We got T as (-at^2,0). SP = \(\sqrt{(a -at)^2 + (2at)^2}\), PQ = \(\sqrt{(at^2 +a)^2 +(2at -2at)^2}\), QT = \(\sqrt{(-a + at^2)^2 +(2at)^2} \), ST =\(\sqrt{(a + at^2)^2+(0)^2}\)
anonymous
  • anonymous
Oh Ok. Thank you! :) I'll post the second part up in a while.
apoorvk
  • apoorvk
i think one part of the above question, where SP=PQ is to be done, can be done straight away by seeing the fact that the basic principal of any point on a parabola is that it is equidistant from the directrix and the focus. (and here we know S(a,0) will be the focus and Q(-a,2at) lies on the directrix. So by the basic definition of the conic, SP=PQ by default. |dw:1332501144510:dw|
apoorvk
  • apoorvk
(sorry forgot to mention why PQ is perpendicular to the directrix (and hence the shortest distance between P and the directrix). because P can have an ordinate '2at'.. same as ordinate of T(-a,2at)

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