# Parabola/Normals

• Jul 7th 2006, 01:02 AM
nath_quam
Parabola/Normals
P(2ap,ap^2) is a point on the parabola x^2=4ay

The normal at P cuts the x axis at S and the y axis at T.

S= (ap(2+p^2),0)
T= (0, a(2+p^2))

Find the value(s) of p such that P is the midpoint of ST

Thanks Nath
• Jul 7th 2006, 05:43 AM
malaygoel
Quote:

Originally Posted by nath_quam
P(2ap,ap^2) is a point on the parabola x^2=4ay

The normal at P cuts the x axis at S and the y axis at T.

S= (ap(2+p^2),0)
T= (0, a(2+p^2))

Find the value(s) of p such that P is the midpoint of ST

Thanks Nath

Is it $\displaystyle p^2 = 2?$
In other words,$\displaystyle p = \sqrt{2}$or$\displaystyle p=-\sqrt{2}$

Keep Smiling
Malay
• Jul 7th 2006, 08:17 AM
Soroban
Hello, Nath!

They already did the groundwork for us . . . the rest is easy.

[I got the same answers, Malay!]

Quote:

$\displaystyle P(2ap,ap^2)$ is a point on the parabola $\displaystyle x^2 = 4ay$

The normal at $\displaystyle P$ cuts the x-axis at $\displaystyle S$ and the y-axis at $\displaystyle T.$

$\displaystyle S \,= \,\left(ap[2+p^2],\,0\right)\qquad T\,=\,\left(0,\,a[2+p^2]\right)$

Find the value(s) of $\displaystyle p$ such that $\displaystyle P$ is the midpoint of $\displaystyle ST.$

I assume you know the Midpoint Formula . . .

Given two points $\displaystyle A(x_1,y_1)$ and $\displaystyle B(x_2,y_2)$, the midpoint of $\displaystyle \overline{AB}$ is: .$\displaystyle \left(\frac{x_1+x_2}{2}\,,\,\frac{y_1+y_2}{2} \right)$

We have: .$\displaystyle S\left(ap[2+p^2],\,0\right),\;\;T\left(0,\,a[2+p^2]\right)$

The midpoint is: .$\displaystyle M \:=\:\left(\frac{ap[2+p^2] + 0}{2}\,,\,\frac{0+a[2+p^2]}{2}\right) \:=$ $\displaystyle \:\left(\frac{ap[2 + p^2]}{2}\,,\,\frac{a[2+p^2]}{2}\right)$

Since we want $\displaystyle M = P$, we have: .$\displaystyle \left(\frac{ap[2+p^2]}{2}\,,\,\frac{a[2+p^2]}{2}\right)\;=\;\left(2ap,\,ap^2\right)$

If two points are equal, their corresponding coordinates are equal.

Equate $\displaystyle x$'s: .$\displaystyle \frac{ap(2+p^2)}{2} = 2ap\quad\Rightarrow\quad ap(2 + p^2) = 4ap$

. . Divide by $\displaystyle ap:\;\;2+p^2 \,=\,4\quad\Rightarrow\quad p^2 = 2\quad\Rightarrow\quad p = \pm\sqrt{2}$

Equate $\displaystyle y$'s: .$\displaystyle \frac{a(2+p^2)}{2} = ap^2\quad\Rightarrow\quad a(2+p^2) = 2ap^2$

. . Divide by $\displaystyle a:\;\;2 + p^2 \,=\,2p^2\quad\Rightarrow\quad p^2 = 2\quad\Rightarrow\quad p = \pm\sqrt{2}$

Therefore: .$\displaystyle \boxed{p \:= \:\pm\sqrt{2}}$
• Jul 7th 2006, 03:56 PM
nath_quam
Thanks Guys
• Jul 10th 2006, 03:39 AM
nath_quam
Just Quickly
Is a possible answer p = 0
• Jul 10th 2006, 03:43 AM
malaygoel
Quote:

Originally Posted by nath_quam
Is a possible answer p = 0

No. It does not satisfy the conditions.

Keep Smiling
Malay