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 $