Q5. The curve \( C \) has equation \[ y=9-4 x-\frac{8}{x}, \quad x>0 \] The point \( P \) on \( C \) has \( x \)-coordinate equal to 2 . (a) Show that the equation of the tangent to \( C \) at the point \( P \) is \( y=1-2 x \). (b) Find an equation of the normal to \( C \) at the point \( P \). The tangent at \( P \) meets the \( x \)-axis at \( A \) and the normal at \( P \) meets the \( x \)-axis at \( B \). (c) Find the area of triangle \( A P B \).

To solve the problem, we will go through each part step by step. ### Part (a) First, we need to find the coordinates of the point \( P \) on the curve \( C \) when \( x = 2 \). The equation of the curve is given by: \[ y = 9 - 4x - \frac{8}{x} \] Substituting \( x = 2 \): \[ y = 9 - 4(2) - \frac{8}{2} = 9 - 8 - 4 = -3 \] Thus, the coordinates of point \( P \) are \( (2, -3) \). Next, we need to find the derivative \( \frac{dy}{dx} \) to determine the slope of the tangent line at point \( P \). Differentiating \( y \): \[ \frac{dy}{dx} = -4 + \frac{8}{x^2} \] Now, substituting \( x = 2 \) into the derivative: \[ \frac{dy}{dx} \bigg|_{x=2} = -4 + \frac{8}{2^2} = -4 + \frac{8}{4} = -4 + 2 = -2 \] The slope of the tangent line at point \( P \) is \( -2 \). Using the point-slope form of the equation of a line, the equation of the tangent line at point \( P(2, -3) \) is: \[ y - (-3) = -2(x - 2) \] Simplifying this: \[ y + 3 = -2x + 4 \implies y = -2x + 1 \] Thus, the equation of the tangent line is: \[ y = 1 - 2x \] This confirms the result. ### Part (b) Next, we need to find the equation of the normal line at point \( P \). The slope of the normal line is the negative reciprocal of the slope of the tangent line. Since the slope of the tangent line is \( -2 \), the slope of the normal line is: \[ \text{slope of normal} = \frac{1}{2} \] Using the point-slope form again for the normal line at point \( P(2, -3) \): \[ y - (-3) = \frac{1}{2}(x - 2) \] Simplifying this: \[ y + 3 = \frac{1}{2}x - 1 \implies y = \frac{1}{2}x - 4 \] Thus, the equation of the normal line is: \[ y = \frac{1}{2}x - 4 \] ### Part (c) Now we need to find the points where the tangent and normal lines intersect the \( x \)-axis. 1. **Finding point \( A \)** (where the tangent meets the \( x \)-axis): Set \( y = 0 \) in the tangent equation: \[ 0 = 1 - 2x \implies 2x = 1 \implies x = \frac{1}{2} \] Thus, point \( A \) is \( \left( \frac{1}{2}, 0 \right) \). 2. **Finding point \( B \)** (where the normal meets the \( x \)-axis): Set \( y = 0 \) in the normal equation: \[ 0 = \frac{1}{2}x - 4 \implies \frac{1}{2}x = 4 \implies x = 8 \] Thus, point \( B \) is \( (8, 0) \). Now we have the vertices of triangle \( APB \): - \( A \left( \frac{1}{2}, 0 \right) \) - \( P(2, -3) \) - \( B(8, 0) \) To find the area of triangle \( APB \), we can use the formula for the area of a triangle given by vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \): \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substituting the coordinates: \[