Parabola/Normals

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

Keep Smiling
Malay
July 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:

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

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

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

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

I assume you know the Midpoint Formula . . .

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

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

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

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

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

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

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

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

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

Therefore: . $\boxed{p \:= \:\pm\sqrt{2}}$
July 7th 2006, 03:56 PM
nath_quam

Thanks Guys
July 10th 2006, 03:39 AM
nath_quam

Just Quickly

Is a possible answer p = 0
July 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