Question

Let the quadratic curve passing through the point \( (-1, 0) \) and touching the line \( y = x \) at \( (1, 1) \) be \( y = f(x) \). Then the x-intercept of the normal to the curve at the point \( (\alpha, \alpha + 1) \) in the first quadrant is:

Answer 1

Correct Answer: 11

Step 1: Equation of the quadratic curve
The general form of the quadratic equation is given by: \[ f(x) = (x + 1)(ax + b) \] We are given that the curve passes through the point \( (-1, 0) \). Substituting \( x = -1 \) and \( f(-1) = 0 \) into the equation: \[ f(-1) = (-1 + 1)(a(-1) + b) = 0 \quad \Rightarrow \quad 1 \cdot (-a + b) = 0 \quad \Rightarrow \quad b = a \] Thus, the equation becomes: \[ f(x) = (x + 1)(ax + a) = a(x + 1)^2 \] Step 2: Deriving the first derivative
The first derivative of \( f(x) \) is: \[ f'(x) = a \cdot 2(x + 1) \] We are also given that the curve touches the line \( y = x \) at \( (1, 1) \), so at this point, the slope of the curve equals the slope of the line \( y = x \), which is 1. Substituting \( x = 1 \) into \( f'(x) \), we get: \[ f'(1) = 2a(1 + 1) = 2a \cdot 2 = 4a \] Equating this to the slope of the line \( y = x \), which is 1: \[ 4a = 1 \quad \Rightarrow \quad a = \frac{1}{4} \] Step 3: Equation of the quadratic curve
Thus, the equation of the curve becomes: \[ f(x) = \frac{1}{4}(x + 1)^2 \] Step 4: Finding the x-coordinate of the point \( (\alpha, \alpha + 1) \)
Now, we need to find the x-intercept of the normal at the point \( (\alpha, \alpha + 1) \). The normal at a point on a curve is perpendicular to the tangent. The slope of the tangent at the point \( (\alpha, \alpha + 1) \) is given by: \[ f'(\alpha) = \frac{1}{2}(\alpha + 1) \] Thus, the slope of the normal is the negative reciprocal: \[ \text{Slope of normal} = -\frac{2}{\alpha + 1} \] Using the point-slope form of the equation of the normal, we get: \[ y - (\alpha + 1) = -\frac{2}{\alpha + 1}(x - \alpha) \] To find the x-intercept, set \( y = 0 \) and solve for \( x \): \[ 0 - (\alpha + 1) = -\frac{2}{\alpha + 1}(x - \alpha) \] \[ -\alpha - 1 = -\frac{2}{\alpha + 1}(x - \alpha) \] \[ \alpha + 1 = \frac{2}{\alpha + 1}(x - \alpha) \] \[ (\alpha + 1)^2 = 2(x - \alpha) \] \[ x = \frac{(\alpha + 1)^2}{2} + \alpha \] Substitute \( \alpha = 3 \) (from previous calculations): \[ x = \frac{(3 + 1)^2}{2} + 3 = \frac{16}{2} + 3 = 8 + 3 = 11 \] Thus, the x-intercept of the normal is \( 11 \).