Step 1:

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The equation is $\"\"$ and the point is $\"\"$ .

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Consider $\"\"$ .

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Differentiate on each side with respect to $\"\"$ .

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$\"\"$

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$\"\"$

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Slope of the tangent is derivetive of the function at the given point.

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$\"\"$

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$\"\"$ .

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Slope of the tangent line is $\"\"$ .

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Step 2:

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Point slope form of line equation is $\"image\"$ .

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Substitute $\"\"$ and $\"\"$ in the above equation.

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$\"\"$

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$\"\"$

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Tangent line is $\"\"$ .

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Normal line is perpendicular to tangent line then \ \

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slope of tangent line $\"\"$ slope of normal line is equal to $\"\"$ .

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Let slope of the normal as $\"\"$ .

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Therefore,

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$\"\"$

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Point slope form of line equation is $\"image\"$ .

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Substitute $\"\"$ and $\"\"$ in the above equation.

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$\"\"$

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$\"\"$

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Normal line equation is $\"\"$ .

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Solution:

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Tangent line is $\"\"$ .

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Normal line equation is $\"\"$ .