Question

If all the normals drawn to the curve \( y = \frac{1 + 3x^2}{3 + x^2} \) at the points of intersection of \( y = \frac{1 + 3x^2}{3 + x^2} \) and \( y = 1 \) pass through the point \( (\alpha, \beta) \), then \( 3\alpha + 2\beta = \):

• A. \( 4 \)
• B. \( 2 \)
• C. \( -2 \)
• D. \( -4 \)

Hint:

The intersection of normals at different points gives the coordinates of the point through which all normals pass.

Answer 1

The Correct Option is A

Step 1: Find the points of intersection.
Set \( y = 1 \): \( 1 = \frac{1 + 3x^2}{3 + x^2} \implies x = \pm 1 \). Points are \( (1, 1) \) and \( (-1, 1) \).
Step 2: Find the derivative \( \frac{dy}{dx} \).
\( \frac{dy}{dx} = \frac{16x}{(3 + x^2)^2} \).
Step 3: Find the slopes of the normals.
At \( (1, 1) \), tangent slope \( = 1 \), normal slope \( = -1 \). Normal equation: \( y - 1 = -1(x - 1) \implies x + y = 2 \).
At \( (-1, 1) \), tangent slope \( = -1 \), normal slope \( = 1 \). Normal equation: \( y - 1 = 1(x + 1) \implies x - y = -2 \).

Step 4: Find the intersection point \( (\alpha, \beta) \). Solving \( x + y = 2 \) and \( x - y = -2 \), we get \( \alpha = 0 \) and \( \beta = 2 \).
Step 5: Calculate \( 3\alpha + 2\beta \).
\( 3(0) + 2(2) = 4 \).