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प्रश्न
Find the equations of tangent and normal to the curve y = 3x2 - 3x - 5 where the tangent is parallel to the line 3x − y + 1 = 0.
उत्तर
Equation of the curve is y = 3x2 - 3x - 5
Differentiating w.r.t. x, we get
`"dy"/"dx"` = 6x - 3
Slope of the tangent at P(x1, y1) is
`("dy"/"dx")_(("x"_1,"x"_2)` = 6x1 - 3
According to the given condition, the tangent is parallel to 3x – y + 1 = 0
Now, slope of the line 3x – y + 1 = 0 is 3.
∴ Slope of the tangent = `"dy"/"dx" = 3`
∴ 6x1 - 3 = 3
∴ x1 = 1
P(x1, y1) lies on the curve y = 3x2 - 3x - 5
∴ y1 = 3(1)2 - 3(1) - 5
∴ y1 = - 5
∴ The point on the curve is (1, -5).
∴ Equation of the tangent at (1, -5) is
∴ (y + 5) = 3(x – 1)
∴ y + 5 = 3x – 3
∴ 3x - y - 8 = 0
Slope of the normal at (1, -5) is `(-1)/("dy"/"dx")_((1,-5)` = `(- 1)/3`
∴ Equation of the normal of (1, -5) is
(y + 5) = `(-1)/3`(x - 1)
∴ 3y + 15 = - x + 1
∴ x + 3y + 14 = 0
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