We know the axis of symmetry for the general quadratic:
$$f(x)=ax^2+bx+c$$
is the line:
$$x=-\frac{b}{2a}$$
And so for the given function:
$$y(x)=5x^2+ax+b$$
The axis of symmetry is:
$$x=-\frac{a}{10}$$
And so the minimum value of $y$ will be:
$$y_{\min}=y\left(-\frac{a}{10}\right)=5\left(-\frac{a}{10}\right)^2+a\left(-\frac{a}{10}\right)+b=\frac{20b-a^2}{20}$$
Now, we are given two points on the parabola, and using this data, we obtain:
$$5a^2+a(a)+b=b$$
$$5b^2+ab+b=a$$
Bearing in mind that $a\ne b$, can you obtain a unique solution to the above system?