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प्रश्न
Find the equation of tangent and normal to the curve at the given points on it.
y = 3x2 - x + 1 at (1, 3)
उत्तर
Equation of the curve is y = 3x2 - x + 1
Differentiating w.r.t. x, we get
`"dy"/"dx"` = 6x - 1
Slope of the tangent at (1, 3) is
`("dy"/"dx")_((1,3)` = 6(1) - 1 = 5
∴ Equation of tangent at (a, b) is
y - b = `("dy"/"dx")_(("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)/(("dy"/"dx")_((1,3)` `= (- 1)/5`
∴ Equation of normal at (a, b) is
y - b = `(-1)/(("dy"/"dx")_(("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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