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

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Aicharya · Accepted Answer

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

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

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