Question

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

• A. $p=2,q=-7$
• B. $p=-2,q=7$
• C. $p=-2,q=-7$
• D. $p=2,q=7$

Answer 1

The correct option is A

$p=2,q=-7$

Explanation for the correct option:

Finding the value of $p$ and $q$.

Given that,

$y^2=p{x}^3+q...\left(1\right)$

Differentiate the given curve with respect to $x$, we get the slope

$$\begin{gathered} 2y\frac{dy}{dx}=3p{x}^2+0 \\ \Rightarrow \frac{dy}{dx}=\frac{3p{x}^2}{2y} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(2,3\right)}=\frac{3p{\left(2\right)}^2}{2\left(3\right)} \\ =2p...\left(2\right) \end{gathered}$$

Since, $y=4x-5$ is tangent to the given curve.

So, slope $m=4$

By comparing the equation $\left(2\right)$ with the above slope, we get

$p=2$.

Substitute the value of $x,y$ and $p$ in the equation $\left(1\right)$, we get the value of $q$

$$\begin{gathered} 3^2=2{\left(2\right)}^3+q \\ \Rightarrow q=9-16 \\ =-7 \end{gathered}$$

Hence, the correct option is A.