c2 differentiation question help please

Sorry i still dont understand how did you get b=-a/10

i got $b = \frac{a^{2}}{20}$
let f(x)=0
$\frac{5a^{2}}{100} -\frac{a^{2}}{10}+b = 0$

$\frac{5a^{2}}{100} -\frac{a^{2}}{10}=-b$

$-(\frac{5a^{2}}{100} -\frac{a^{2}}{10}) = b$

$\frac{a^{2}}{10} -\frac{5a^{2}}{100} =b$


where am i going wrong

Why have you set f(x) = 0? That tells you where f(x) crosses the axis, not where it has a turning point!

You showed that at a turning point you need x = -a/10. However, the question tells you that the turning point is at (b, a) i.e. x = b and y = a.

So this tells you 2 things:

x = b and x = -a/10 are the same coordinate i.e. b = -a/10

and

f(b) = a because the point (b, a) lies on the curve.

Can you solve from there?