If \(y=4 x-5\) is a tangent to the curve \(y^2=\mathrm{p} x^3+\mathrm{q}\) at \((2,3)\), then \(\mathrm{p}-\mathrm{q}\) is

\(y^2=\mathrm{p} x^3+\mathrm{q}... (i)\)
Differentiating both sides w.r.t. \(x\), we get
\(\begin{gathered}
2 y \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=3 \mathrm{p} x^2 \\
\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 \mathrm{p}}{2}\left(\frac{x^2}{y}\right) \\
\therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(2,3)}=\frac{3 \mathrm{p}}{2} \times \frac{4}{3}=2 \mathrm{p}
\end{gathered}\)
Slope of the line \(y=4 x-5\) is 4 .
Since the line touches the curve, their slopes are equal.
\(\therefore \quad 2 \mathrm{p}=4 \Rightarrow \mathrm{p}=2\)
Since \((2,3)\) lies on \(y^2=\mathrm{p} x^3+\mathrm{q}\).
\(\begin{array}{ll}
\therefore & 9=2 \times 8+q \Rightarrow q=-7 \\
\therefore & p-q=2+7=9
\end{array}\)