Given a parabolic shape with maximum height $OC=8m$ and maximum width $AB=20m$. If $M$ is the middle of $OB$, then what is the height $MK$, from $M$?

Question

Given a parabolic shape with maximum height $OC=8m$ and maximum width $AB=20m$. If $M$ is the middle of $OB$, then what is the height $MK$, from $M$?

I attempted to solve the question as follows:

$OB=10m$

Hence, $OM=5m$

$OC=8m$

I state that $O$ is point $(0,0)$ on the Cartesian coordinate system.

$\implies M(5, 0), C(0, 8), B(0, 10), A(-10, 0)$. I then attempted to find the coordinates of K, but I don't know how to. Obviously, $\triangle AKB$ is isosceles, but I couldn't see anything else further than that. Could you please explain to me how to solve this question?

Answer 1

From A(-10,0)and B(10,0), x=-10 and x=10 are two roots:

\begin{align}y(x)=a(x+10)(x-10)\end{align}

From C(0,8), a= -$\frac{8}{100}$

\begin{align}y(x) =-\frac{8}{100}(x+10)(x-10)\end{align} \begin{align}y(5) =-\frac{8}{100}(5+10)(5-10)=6 \end{align}

K's coordinates (5 cm,6 cm)

Answer 2

Turning Point form of equation for parabola is $y=a(x-h)^2+k$

$C$ is TP, so $h=0, k=8$.

$A$ is at $(-10,0)$, subst for $x, y$

So $0=a(0-10)^2+8 \rightarrow a=-\frac {2} {25}$

$M$ is at $x=5 \rightarrow y=-\frac {2} {25}5^2+8$

Height $MK$ is 6m

Answer 3

For the given question, you can simply solve using basic properties of parabola and without using coordinate geometry.

Draw a perpendicular to $CO$ from point $K$ and say it meets $CO$ at $H$.

We observe that axis of the parabola is along $OC$ and $C$ is the vertex of the parabola. Second, we know that the perpendicular distance from the vertex to an ordinate is directly proportional to the square of the ordinate. So we have,

$CO = k \cdot OB^2$ and $CH = k \cdot HK^2$ where $k$ is some constant.

As $ \ CH = CO - HO = CO - MK$ and $HK = OM$,

$\displaystyle \frac{CO}{CO - MK} = \frac{OB^2}{OM^2} = 4 \ $ (as $OB = 2 \cdot OM$)

Since $CO = 8, \ MK = 6$