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
Hello, Nath!
They already did the groundwork for us . . . the rest is easy.
[I got the same answers, Malay!]
$\displaystyle P(2ap,ap^2)$ 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: .$\displaystyle \boxed{p \:= \:\pm\sqrt{2}}$