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प्रश्न
Find the equations of tangent and normal to the curve y = 3x2 – x + 1 at the point (1, 3) on it
उत्तर
Equation of the curve is y = 3x2 – 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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