Find the equation of normal to the curve y = x-3 which is perpendicular to the line 6x + 3y – 4 = 0. - Mathematics and Statistics

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Sum

Find the equation of normal to the curve y = `sqrt(x - 3)` which is perpendicular to the line 6x + 3y – 4 = 0.

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Solution

Equation of the curve is y = `sqrt(x - 3)`

Differentiating w.r.t. x, we get

`"dy"/"dx" = 1/(2sqrt("x - 3"))`

Slope of the tangent at P(x1, y1) is

`(("d"y)/("d"x))_((x_1","  y_1)) = 1/(2sqrt(x_1 - 3))`

Slope of the line 6x + 3y – 4 = 0 is `(-6)/3` = – 2.

According to the given condition, tangent to the curve is perpendicular to the line 6x + 3y – 4 = 0.

∴ slope of the tangent = `(("d"y)/("d"x))_((x_1, y_1)` = `1/2`

∴ `1/(2sqrt(x_1 - 3)) = 1/2`

∴ `sqrt(x_1 - 3) = 1`

∴ x1 – 3 = 1

∴ x1 = 4

P(x1, y1) lies on the curve y = `sqrt(x - 3)`

∴ `y_1 = sqrt(4 - 3)`

∴ y1 = 1

∴ The require point is (4, - 1) or (4, 1).

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Chapter 4: Applications of Derivatives - Miscellaneous Exercise 4 [Page 114]

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