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: $\boxed{x = 1}$

Answer 2

Step 1: Understand the problem
Given the curve: \((x^2 + 1)(y - 3) = x\).
We need to find the equation of the normal at point \(P\) in the first quadrant where the tangent is horizontal.

Step 2: Differentiate the curve implicitly
Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx} \big[(x^2 + 1)(y - 3)\big] = \frac{d}{dx}(x) \]
Using product rule:
\[ 2x(y - 3) + (x^2 + 1)\frac{dy}{dx} = 1 \]

Step 3: Solve for \(\frac{dy}{dx}\)
\[ (x^2 + 1) \frac{dy}{dx} = 1 - 2x(y - 3) \]
\[ \frac{dy}{dx} = \frac{1 - 2x(y - 3)}{x^2 + 1} \]

Step 4: Use the condition for horizontal tangent
Tangent is horizontal \(\Rightarrow \frac{dy}{dx} = 0\), so:
\[ 1 - 2x(y - 3) = 0 \implies 2x(y - 3) = 1 \implies y - 3 = \frac{1}{2x} \]

Step 5: Use the original curve equation to find \(x\)
Recall from the original curve:
\[ (x^2 + 1)(y - 3) = x \]
Substitute \(y - 3 = \frac{1}{2x}\):
\[ (x^2 + 1) \times \frac{1}{2x} = x \]
Multiply both sides by \(2x\):
\[ x^2 + 1 = 2x^2 \implies 1 = x^2 \]
Since \(P\) lies in the first quadrant, \(x > 0\), so:
\[ x = 1 \]

Step 6: Find corresponding \(y\)
From step 4:
\[ y - 3 = \frac{1}{2 \times 1} = \frac{1}{2} \implies y = \frac{7}{2} = 3.5 \]

Step 7: Equation of normal at \(P=(1, 3.5)\)
Since tangent is horizontal, slope of tangent \(m_t = 0\), so slope of normal \(m_n\) is undefined (vertical line).
Thus, normal is vertical line through \(x = 1\).

Final answer:
\[ \boxed{x = 1} \]