Question

The equation of the normal drawn to the curve \( y^3 = 4x^5 \) at the point \( (4,16) \) is:

• A. \( 20x + 3y = 128 \)
• B. \( 20x - 3y = 32 \)
• C. \( 3x - 20y + 308 = 0 \)
• D. \( 3x + 20y = 332 \) \bigskip

Hint:

For finding the equation of the normal, first find the derivative of the curve to get the slope of the tangent. The slope of the normal is the negative reciprocal of the tangent slope.

Answer 1

The Correct Option is D

We are given the curve equation: \[ y^3 = 4x^5. \] To find the equation of the normal, we first differentiate the given equation with respect to \( x \). Step 1: Differentiate implicitly. \[ \frac{d}{dx} (y^3) = \frac{d}{dx} (4x^5). \] Using the chain rule for \( y^3 \) and power rule for \( 4x^5 \), we get: \[ 3y^2 \frac{dy}{dx} = 20x^4. \] Now solve for \( \frac{dy}{dx} \) (slope of the tangent): \[ \frac{dy}{dx} = \frac{20x^4}{3y^2}. \] At the point \( (4, 16) \), substitute \( x = 4 \) and \( y = 16 \) into the above equation: \[ \frac{dy}{dx} = \frac{20 \cdot 4^4}{3 \cdot 16^2} = \frac{20 \cdot 256}{3 \cdot 256} = \frac{20}{3}. \] Step 2: Slope of the normal. The slope of the normal is the negative reciprocal of the slope of the tangent. Hence, the slope of the normal is: \[ m_{\text{normal}} = -\frac{3}{20}. \] Step 3: Equation of the normal. The equation of the normal to a curve at the point \( (x_1, y_1) \) is given by: \[ y - y_1 = m_{\text{normal}} (x - x_1). \] Substituting \( m_{\text{normal}} = -\frac{3}{20} \), \( x_1 = 4 \), and \( y_1 = 16 \): \[ y - 16 = -\frac{3}{20}(x - 4). \] Simplify the equation: \[ y - 16 = -\frac{3}{20}x + \frac{12}{20}. \] Multiply through by 20 to eliminate the denominator: \[ 20y - 320 = -3x + 12. \] Rearranging terms gives the equation of the normal: \[ 3x + 20y = 332. \] Thus, the correct answer is \( 3x + 20y = 332 \). \bigskip