Question

If the tangent drawn to the curve $(x^2+1)(y-3)=x$ at a point P, lying in the first quadrant, is a horizontal line, then the equation of the normal at the point P is

• A. $x=\dfrac{7}{2}$
• B. $x=1$
• C. $y=\dfrac{7}{2}$
• D. $y=1$

Hint:

Horizontal Tangents and Normals:

• Horizontal tangent: $\fracdydx = 0$
• Normal is perpendicular: If tangent is horizontal, normal is vertical $\Rightarrow x = x_0$
• Always check quadrant restrictions

Answer 1

The Correct Option is B

Given: $(x^2 + 1)(y - 3) = x$. Rewriting: $y = \dfrac{x}{x^2 + 1} + 3$. Differentiate: \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x}{x^2 + 1} \right) = \frac{(x^2 + 1)(1) - x(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \] Set $\frac{dy}{dx} = 0$: \[ \Rightarrow 1 - x^2 = 0 \Rightarrow x = \pm 1 \] First quadrant: $x=1$. Find $y$: \[ y = \frac{1}{1^2 + 1} + 3 = \frac{1}{2} + 3 = \frac{7}{2} \Rightarrow P = \left(1, \frac{7}{2}\right) \] Tangent is horizontal $\Rightarrow$ normal is vertical: $\boxed{x = 1}$