Question

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

• A. $-5$
• B. $5$
• C. $-9$
• D. $9$
• E. $0$

Answer 1

The correct option is A $-5$

Given curve is $y^2=p{x}^3+q....\left(i\right)$
On differentiating w.r.t. $x$, we get
$2y\frac{dy}{dx}=3p{x}^2\Rightarrow \frac{dy}{dx}=\frac{3p{x}^2}{2y}$
At $(2,3),{\left(\frac{dy}{dx}\right)}_{(2,3)}=\frac{12p}{6}=2p$
Since, line $y=4x-5$ is a tangent to the given curve.
Therefore, $\text{slope}=4=2p$
$\Rightarrow p=2$
Now, put the value of $p,x$ and $y$ in Eq. (i), we get
$(3{)}^2=(2\left)\right(2{)}^3+q$
$\Rightarrow 9=16+q$
$\Rightarrow q=-7$
Hence, $p+q=2-7=-5$.