Answer: \(p=2\), \(q=6\).

We start with the main method. Differentiate to obtain the tangent gradient, then use the negative reciprocal for the normal gradient where needed.

The curve is

\(\displaystyle y=\frac{p}{x^2}+5x-2=px^{-2}+5x-2.\)

Differentiate:

\(\frac{\mathrm dy}{\mathrm dx}=-2px^{-3}+5.\)

At \(x=1\), the tangent gradient is

\(-2p+5.\)

The normal has equation \(y=-x+q\), so its gradient is \(-1\).

Therefore the tangent gradient is \(1\).

So

\(-2p+5=1.\)

This gives

\(p=2.\)

Now find the point on the curve when \(x=1\):

\(y=\frac{2}{1^2}+5(1)-2=5.\)

The point is \((1,5)\). Since it lies on \(y=-x+q\),

\(5=-1+q.\)

Therefore

\(q=6.\)

This completes the solution and gives the required result.