Maharashtra State BoardHSC Commerce 12th Board Exam
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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. - Mathematics and Statistics

Sum

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.

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Solution

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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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 4 Applications of Derivatives
Exercise 4.1 | Q 3 | Page 105
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