anonymous
  • anonymous
Need assurance and a little help! A parabola is given parametrically by x=at^2 , y= 2at. If P is any point on the parabola, let F be the foot of the perpendicular from P onto the axis of symmetry. Let G be the point where the normal from P crosses the axis of symmetry. Prove that FG=2a
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 ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Then 2a/2at = 1/t

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anonymous
  • anonymous
Use the equation of Normal \[y - 2at = -\frac{1}{\frac1t}(x - at^2) \]
anonymous
  • anonymous
Which is y=(1/t)x + at
anonymous
  • anonymous
??No
anonymous
  • anonymous
Normal is perpendicular to tangent. \[m_{t} m_n = -1\]
anonymous
  • anonymous
OH, Ok.
anonymous
  • anonymous
Axis of symmetry is the X-axis.
anonymous
  • anonymous
How do you get that in the form y=mx+b?
dumbcow
  • dumbcow
isn't F and G the same since the normal is also perpendicular to P and both are meeting at x-axis ?
anonymous
  • anonymous
I am confused now :/
anonymous
  • anonymous
I'm not sure.. :/
anonymous
  • anonymous
dumbcow's right I think
dumbcow
  • dumbcow
no guys what i really meant was i think theres a typo
dumbcow
  • dumbcow
maybe it should be F is where tangent line meet x-axis ?
anonymous
  • anonymous
That's what it says in the book... (I mean what I wrote)
anonymous
  • anonymous
Normal:\[y - 2at = -xt + at^3\]\[y + xt = at^3 + 2at\]\(y = 0\)\(\implies x = at^2 + 2at\) Tangent: \[y = \frac{1}{t}x +at\]\(y = 0\) \(\implies x = -at^2\)
anonymous
  • anonymous
Why at^3 and not at^2?
anonymous
  • anonymous
\[y - 2at = -t(x - at^2)\]
anonymous
  • anonymous
The equation of Normal
anonymous
  • anonymous
Ok, and how do you carry on from there?
anonymous
  • anonymous
I don't know, the question stops making sense from here.
anonymous
  • anonymous
hmmm Ok....
dumbcow
  • dumbcow
lol

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