Midpoint Property of Tangents to Parabolas

[chwala]

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?

 

[verty]

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 ${Q.y + R.y \over 2}$.

 

[neilparker62]

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.

 

[chwala]

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?

 

[chwala]

$y= 2a(x+x1)/y1$
$y^2=4a(x+b)$
$4yy1=4a(x+x1)$ and $y^2=4a(x+b)$
on subtraction...
$4yy1-y^2=4ax1-4ab$
advise

 

[neilparker62]

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.

 

[Mark44]

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.

 

[neilparker62]

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)

 

[Mark44]

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.

 

[chwala]

so what is the conclusion

 

[neilparker62]

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.

 

[chwala]

ok let me look at it...

 

[chwala]

this is still pending...i need help

 

[Mark44]

@neilparker62 seems to have given you a hint. Have you tried it?

 

[chwala]

let me look at it, i want to clear all pending assignments here...