Proof Help

**arze** · October 18th 2009, 03:49 AM

Show that the equation of the normal to the parabola $y^2=ax$ at the point $(at^2, 2at0$ is $y+tx=2at+at^3$ .
If this normal meets the curve again at the point $(aT^2,2aT)$ show that
$t^2+tT+2=0$
and deduce that $T^2$ is greater than 8.

My problem here is deducing the $T^2$ is greater than 8.
Thanks

**nolanfan** · October 18th 2009, 04:26 AM

Originally Posted by arze

Show that the equation of the normal to the parabola $y^2=ax$ at the point $(at^2, 2at0$ is $y+tx=2at+at^3$ .
If this normal meets the curve again at the point $(aT^2,2aT)$ show that
$t^2+tT+2=0$
and deduce that $T^2$ is greater than 8.

My problem here is deducing the $T^2$ is greater than 8.
Thanks

Its $y^2=4ax$ right?

Well since you've deduced $t^2+Tt+2=0$ it is easy to show that $T^2$ is greater than 8. Since the normal intersects the parabola in 2 points, the discriminant of the equation in t has to be greater than zero (for a tangent, discriminant = 0, discriminant is denoted by delta). Therefore, we get

$\Delta \textgreater0, <br /> T^2 - 4(2) \textgreater0,<br /> T^2 \textgreater8$

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