Question

A deflected beam is the shape of parabola $x^2=4ay$. The beam is supported at its ends which are 12 cm apart. The deflection in the centre is 9 cm only. Find the points on the beam where deflection is 5 cm.

Answer 1

Let PQ=12 m and $OM=9cm=\frac{9}{100}m$

Since, QM=MP=6. therefore the coordinates of point P are $(6,\frac{9}{100})$.

Let the equation of the parabolic beam POQ be $x^2=4ay$.

$\Rightarrow 6^2=4a\left(\frac{9}{100}\right)\Rightarrow a=100$

$\therefore$ Equation is $x^2=400y$

Let deflection $AC=5cm\Rightarrow BC=9-5=4cm$

Hence, coordinates of C can be taken as $(x,\frac{4}{100})$.

$\therefore x^2=400\times \frac{4}{100}=16\Rightarrow x=\pm 4$

Hence, the two possible points where deflection is 5 cm, are $C(4,\frac{1}{25})$ and $C^{\prime }(-4,\frac{1}{25})$.