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Find the equations of tangent and normal to the curve y = 3x^{2} – x + 1 at the point (1, 3) on it

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#### Solution

Equation of the curve is y = 3x^{2} – x + 1

Differentiating w.r.t. x, we get

`("d"y)/("d"x)` = 6x – 1

Slope of the tangent at (1, 3) is

`(("d"y)/("d"x))_((1, 3)` = 6(1) – 1 = 5

∴ Equation of tangent at (a, b) is

y – b = `(("d"y)/("d"x))_(("a""," "b")) (x - "a")`

Here, (a, b) ≡ (1, 3)

∴ Equation of the tangent at (1, 3) is

(y – 3) = 5(x – 1)

∴ y – 3 = 5x – 5

∴ 5x – y – 2 = 0

Slope of the normal at (1, 3) is `(-1)/(("d"y)/("d"x))_((1"," 3)) = (-1)/5`

∴ Equation of normal at (a, b) is

y – b = `(-1)/(("d"y)/("d"x))_(("a""," "b")) (x - "a")`

∴ Equation of the normal at (1, 3) is

(y – 3) = `(-1)/5 (x - 1)`

∴ 5y – 15 = – x + 1

∴ x + 5y – 16 = 0

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