Question

The combined equation of the tangent and normal to the curve \( xy = 15 \) at the point (5, 3) is

• A. \( 15x^2 - 15y^2 + 16xy = 480 \)
• B. \( 15x^2 + 16xy - 198x + 10y + 480 - 15y^2 = 0 \)
• C. \( 15x^2 - 16xy + 19x - 10y - 480 + 15y^2 = 0 \)
• D. \( 3x + 5y - 30 = 0 \) (simplified individual)

Hint:

Joint equation of two lines $L_1$ and $L_2$ is found by $L_1 \cdot L_2 = 0$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the equation of the tangent and the normal separately and then multiply them to get the combined equation.
Step 2: Key Formula or Approach:
1. Find \( \frac{dy}{dx} \) to get the slope of the tangent.
2. Use point-slope form for tangent and normal.
3. Multiply the two linear equations.
Step 3: Detailed Explanation:
Curve: \( xy = 15 \implies y = \frac{15}{x} \).
Differentiating: \( \frac{dy}{dx} = -\frac{15}{x^2} \).
At \( (5, 3) \), slope of tangent \( m_t = -\frac{15}{5^2} = -\frac{15}{25} = -\frac{3}{5} \).
Equation of tangent: \( y - 3 = -\frac{3}{5}(x - 5) \implies 5y - 15 = -3x + 15 \implies 3x + 5y - 30 = 0 \).
Slope of normal \( m_n = \frac{5}{3} \).
Equation of normal: \( y - 3 = \frac{5}{3}(x - 5) \implies 3y - 9 = 5x - 25 \implies 5x - 3y - 16 = 0 \).
Combined equation:
\[ (3x + 5y - 30)(5x - 3y - 16) = 0 \] \[ 15x^2 - 9xy - 48x + 25xy - 15y^2 - 80y - 150x + 90y + 480 = 0 \] \[ 15x^2 + 16xy - 15y^2 - 198x + 10y + 480 = 0 \] Step 4: Final Answer:
The combined equation is \( 15x^2 + 16xy - 198x + 10y + 480 - 15y^2 = 0 \).