Tangent/Parabola question

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chwala
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Homework Statement


[/B]
If the tangent at the point ##P## to the parabola ## y^2=4ax## meets the parabola ## y^2=4a(x+b)## at points ##Q## and ##R##. Prove that ##P## is the midpoint of ##QR##


Homework Equations




The Attempt at a Solution


We know that the tangent equation to the parabola ## y^2=4ax## is given by ##yy1=2a(x+x1)##
and that ## y^2=4a(x+b)## it follows that
##2a(x+x1) = 4a(x+b)##
→## (x+x1)=2(x+b)##
→##P= 2QR##
or
→##1/2P=QR##
is my working correct?
 
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  • #2
verty
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Homework Statement


[/B]
If the tangent at the point ##P## to the parabola ## y^2=4ax## meets the parabola ## y^2=4a(x+b)## at points ##Q## and ##R##. Prove that ##P## is the midpoint of ##QR##


Homework Equations




The Attempt at a Solution


We know that the tangent equation to the parabola ## y^2=4ax## is given by ##yy1=2a(x+x1)##
and that ## y^2=4a(x+b)## it follows that
##2a(x+x1) = 4a(x+b)##
→## (x+x1)=2(x+b)##
→##P= 2QR##
or
→##1/2P=QR##
I don't understand that at all. Can you explain how you get P = 2QR from the stuff above it?

I know of a way to prove this but it is difficult. Start like this, let ##P.x = u##. Find ##Q.x## and ##R.x## in terms of u (not so tough). Then find ##{Q.y + R.y \over 2}##.
 
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  • #3
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yy1 is not the same as y^2. Solve simultaneously y=2a(x+x1)/y1 and y^2=4a(x+b). Add the roots which will eliminate a lot of stuff - and divide by 2. With any luck you will end up with x1.
 
  • #4
chwala
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yy1 is not the same as y^2. Solve simultaneously y=2a(x+x1)/y1 and y^2=4a(x+b). Add the roots which will eliminate a lot of stuff - and divide by 2. With any luck you will end up with x1.
how?
 
  • #5
chwala
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##y= 2a(x+x1)/y1##
##y^2=4a(x+b)##
##4yy1=4a(x+x1)## and ##y^2=4a(x+b)##
on subtraction...
##4yy1-y^2=4ax1-4ab##
advise
 
  • #6
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You need to square the first equation so that the LHS in both equations is y^2. Then: 4a(x+b)=4a^2(x+x1)^2/(y1)^2.
 
  • #7
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I don't understand that at all.
I agree.

Can you explain how you get P = 2QR from the stuff above it?
As already stated, this makes no sense. P, Q, and R are the names of the three points in this problem -- they aren't numbers. It would be better to identify the x and y coordinates at each point something like this:
##P(x_p, y_p), Q(x_q, y_q), R(x_r, y_r)##

We know that the tangent equation to the parabola ## y^2=4ax## is given by ##yy1=2a(x+x1)##
I don't follow this at all. The slope of the tangent line at point P is ##\frac{2a}{y_p}##. The equation of the tangent line at point P is ##y - y_p = \frac{2a}{y_p}(x - x_p)##, using the point-slope form of the line equation and thenotation I described above.
 
  • #8
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I don't follow this at all. The slope of the tangent line at point P is ##\frac{2a}{y_p}##. The equation of the tangent line at point P is ##y - y_p = \frac{2a}{y_p}(x - x_p)##, using the point-slope form of the line equation and thenotation I described above.
Because you will multiply through by y_p obtaining a term (y_p)^2=4a(x_p)
 
  • #9
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Because you will multiply through by y_p obtaining a term (y_p)^2=4a(x_p)
OK, I see it now, but it wasn't immediately obvious.
 
  • #10
chwala
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so what is the conclusion
 
  • #11
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Solve simultaneously y=2a(x+x1)/y1 and y^2=4a(x+b). Add the roots which will eliminate a lot of stuff - and divide by 2. With any luck you will end up with x1.
 
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  • #12
chwala
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ok let me look at it...
 
  • #13
chwala
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this is still pending...i need help
 
  • #15
chwala
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let me look at it, i want to clear all pending assignments here...
 

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