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The slope of the normal to the curve y=2x2y = 2x^2 at x=1x = 1 is:

Solution

✅ Correct Option: 4

Given: y=2x2y = 2x^2 at x=1x = 1

To find the slope of the normal line, first find the slope of the tangent line using the derivative.

dydx=4x\frac{dy}{dx} = 4x


At x=1x = 1:

dydx∣x=1=4(1)\frac{dy}{dx}\bigg|_{x=1} = 4(1)

dydx∣x=1=4\frac{dy}{dx}\bigg|_{x=1} = 4

The slope of the tangent line is 44.


The normal line is perpendicular to the tangent line.

For perpendicular lines, the slopes are negative reciprocals:

Slope of Normal=−1Slope of Tangent\text{Slope of Normal} = -\frac{1}{\text{Slope of Tangent}}

Slope of Normal=−14\text{Slope of Normal} = -\frac{1}{4}


Therefore, the slope of the normal to the curve at x=1x = 1 is −14-\frac{1}{4}.

Note: The slope of the tangent is 44, while the slope of the normal is −14-\frac{1}{4}.

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