Question

\(P(5,2)\) is a point on the curve \(y=f(x)\) and \(\frac{7}{2}\) is the slope of the tangent to the curve at P. The area of the triangle formed by the tangent and the normal to the curve at P with x-axis is

• A. 35
• B. \(\frac{35}{2}\)
• C. \(\frac{53}{7}\)
• D. \(\frac{53}{14}\)

Hint:

Area of triangle formed by tangent, normal, and x-axis is given by \(\frac{1}{2} |y_P| |x_N - x_T|\). Or using formula: Area = \(\frac{y^2 |1 + m^2|}{2|m|}\) where \(m\) is slope of tangent.

Answer 1

The Correct Option is C

Step 1: Tangent Equation:
Point \(P(5, 2)\), Slope \(m_T = \frac{7}{2}\). Equation: \(y - 2 = \frac{7}{2}(x - 5)\). X-intercept of Tangent (\(A\)): Set \(y=0\). \(-2 = \frac{7}{2}(x - 5) \implies -\frac{4}{7} = x - 5 \implies x_A = 5 - \frac{4}{7} = \frac{31}{7}\).
Step 2: Normal Equation:
Slope of Normal \(m_N = -\frac{1}{m_T} = -\frac{2}{7}\). Equation: \(y - 2 = -\frac{2}{7}(x - 5)\). X-intercept of Normal (\(B\)): Set \(y=0\). \(-2 = -\frac{2}{7}(x - 5) \implies 7 = x - 5 \implies x_B = 12\).
Step 3: Calculate Area:
Vertices are \(P(5,2)\), \(A(\frac{31}{7}, 0)\), \(B(12, 0)\). Base on x-axis = \(|x_B - x_A| = |12 - \frac{31}{7}| = \frac{84 - 31}{7} = \frac{53}{7}\). Height = y-coordinate of P = 2. Area = \(\frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times \frac{53}{7} \times 2 = \frac{53}{7}\).