What is the equation of the normal line to y = x^2 at the point (2, 4)?
Answer and explanation
Correct answer: C. x + 4y - 18 = 0
- Step 1: Find the tangent slope: dy/dx = 2x. At x = 2, m_tan = 2(2) = 4.
- Step 2: The normal line is perpendicular to the tangent, so m_norm = -1/4.
- Step 3: Use point-slope form with (2, 4): y - 4 = -1/4(x - 2).
- Step 4: Multiply by 4: 4y - 16 = -x + 2, which rearranges to x + 4y - 18 = 0.
Answer
x + 4y - 18 = 0.
Why not the others: '4x - y - 4 = 0' gives the tangent line equation rather than the normal line; 'x - 4y + 14 = 0' uses a positive slope +1/4; '4x + y - 12 = 0' inverts coordinates in the line equation.
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