Question

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

Answer 1

x' = 2at
y'= 2a ?

Answer 2

Yes

Answer 3

Then 2a/2at = 1/t

Answer 4

Use the equation of Normal
\[y - 2at = -\frac{1}{\frac1t}(x - at^2) \]

Answer 5

Which is
y=(1/t)x + at

Answer 6

??No

Answer 7

Normal is perpendicular to tangent.
\[m_{t} m_n = -1\]

Answer 8

OH, Ok.

Answer 9

Axis of symmetry is the X-axis.

Answer 10

How do you get that in the form y=mx+b?

Answer 11

isn't F and G the same since the normal is also perpendicular to P and both are meeting at x-axis ?

Answer 12

I am confused now :/

Answer 13

I'm not sure.. :/

Answer 14

dumbcow's right I think

Answer 15

no guys what i really meant was i think theres a typo

Answer 16

maybe it should be F is where tangent line meet x-axis ?

Answer 17

That's what it says in the book... (I mean what I wrote)

Answer 18

Normal:\[y - 2at = -xt + at^3\]\[y + xt = at^3 + 2at\]\(y = 0\)\(\implies x = at^2 + 2at\)
Tangent: \[y = \frac{1}{t}x +at\]\(y = 0\) \(\implies x = -at^2\)

Answer 19

Why at^3 and not at^2?

Answer 20

\[y - 2at = -t(x - at^2)\]

Answer 21

The equation of Normal

Answer 22

Ok, and how do you carry on from there?

Answer 23

I don't know, the question stops making sense from here.

Answer 24

hmmm Ok....

Answer 25

lol