Originally Posted by

**Shiraz** From (30,20):

$\displaystyle 20 = a(30-p)^2$

Also, at (30,20), the gradients of the two parabolas are equal, since the graph joins 'smoothly'.

For $\displaystyle y = -\frac{2}{45}x^2+60$ ,$\displaystyle \frac{dy}{dx} = -\frac{4}{45}x$, so at x = 30 the gradient is $\displaystyle -\frac{8}{3}$.

For $\displaystyle y = a(x-p)^2$, $\displaystyle \frac{dy}{dx} = 2a(x-p)$, so at x = 30 the gradient is $\displaystyle 2a(30-p)$

So$\displaystyle -\frac{8}{3} = 2a(30-p)$

What do we do next? We still have two unknowns? How can we solve for a and p?