Question

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

• A. $p=2,q=-7$
• B. $p=-2,q=7$
• C. $p=-2,q=-7$
• D. $p=2,q=7$
• E. $p=0,q=7$

Answer 1

Answer: (A)

The equation of curve is ${{y}^{2}}=p{{x}^{3}}+q$
$\therefore$ $2y\frac{dy}{dx}=\frac{3p{{x}^{2}}}{2y}$
$\Rightarrow$ $\frac{dy}{dx}=\frac{3p{{x}^{2}}}{2y}$
$\therefore$ ${{\left( \frac{dy}{dx} \right)}_{(2,3)}}=\frac{3p{{(2)}^{2}}}{2.3}=2p$
The equation of tangent at (2, 3) is $(y-3)=2p(x-2)$
$\Rightarrow$ $2px-y=4p-3$ ...(i)
This is similar to $y=4x-5$
$\therefore$ $2p=4$ and $4p-3=5$
$\Rightarrow$ $p=2$ and $p=2$
The point (2, 3) lies on the curve.
$\therefore$ $9=8p+q$
$\Rightarrow$ $9=16+q$ ( $\because$ $p=2$ )
$\Rightarrow$ $q=-7$