Question

The equation of the normal to the curve \( y = \cosh x \) drawn at the point nearest to the origin is:

• A. \( y = 0 \)
• B. \( x = 1 \)
• C. \( x = 0 \)
• D. \( y = 1 \)

Hint:

To find the normal to a curve at a point, compute the slope of the tangent using the derivative, then use the perpendicular slope to write the normal’s equation.

Answer 1

The Correct Option is C

Step 1: Find the point on the curve \( y = \cosh x \) nearest to the origin.
The curve is \( y = \cosh x \), where \( \cosh x = \frac{e^x + e^{-x}}{2} \), and the origin is \((0, 0)\). The distance from a point \((x, y)\) on the curve to the origin is: \[ d = \sqrt{x^2 + y^2} = \sqrt{x^2 + (\cosh x)^2} \] To minimize the distance, minimize \( d^2 = x^2 + (\cosh x)^2 \). Let \( f(x) = x^2 + (\cosh x)^2 \). Compute the derivative: \[ (\cosh x)^2 = \left( \frac{e^x + e^{-x}}{2} \right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4} = \frac{\cosh 2x + 1}{2} \] So: \[ f(x) = x^2 + \frac{\cosh 2x + 1}{2} \] \[ f'(x) = 2x + \frac{1}{2} \cdot \sinh 2x \cdot 2 = 2x + \sinh 2x \] Set \( f'(x) = 0 \): \[ 2x + \sinh 2x = 0 \implies \sinh 2x = -2x \] Since \( \sinh 2x = 2 \sinh x \cosh x \), this is a transcendental equation. Test \( x = 0 \): \[ \sinh 2(0) = \sinh 0 = 0, \quad -2(0) = 0 \implies 0 = 0 \] So, \( x = 0 \) is a critical point. Check the second derivative to confirm a minimum: \[ f''(x) = 2 + \cosh 2x \cdot 2 = 2 + 2 \cosh 2x \] At \( x = 0 \): \[ f''(0) = 2 + 2 \cosh 0 = 2 + 2 \cdot 1 = 4>0 \] This indicates a minimum. At \( x = 0 \), \( y = \cosh 0 = 1 \), so the point is \((0, 1)\), with distance \( d = \sqrt{0 + 1^2} = 1 \). Other points have larger distances (e.g., at \( x = 1 \), \( y = \cosh 1 \approx 1.543 \), \( d \approx \sqrt{1 + (1.543)^2} \approx 1.836 \)), so \((0, 1)\) is indeed the nearest.
Step 2: Find the equation of the normal at \((0, 1)\).
The slope of the tangent to \( y = \cosh x \) is: \[ \frac{dy}{dx} = \sinh x \] At \( x = 0 \): \[ \text{Slope of tangent} = \sinh 0 = 0 \] The tangent at \((0, 1)\) is horizontal (\( y = 1 \)). The normal is perpendicular to the tangent, so its slope is undefined (vertical line). The equation of the normal passing through \((0, 1)\) is: \[ x = 0 \] This matches option (3). Final Answer: \[ \boxed{3} \]