If $y=4 x-5$ is a tangent to the curve $y^2=\mathrm{p} x^3+\mathrm{q}$ at $(2,3)$, then…
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
-5
5
9
-9
Solution
$y^2=\mathrm{p} x^3+\mathrm{q}... (i)$
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 $y^2=\mathrm{p} x^3+\mathrm{q}$.
$\begin{array}{ll}
\therefore & 9=2 \times 8+q \Rightarrow q=-7 \\
\therefore & p-q=2+7=9
\end{array}$