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

Sum

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

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).

Is there an error in this question or solution?
Chapter 4: Applications of Derivatives - Miscellaneous Exercise 4 [Page 114]

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Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 4 Applications of Derivatives
Miscellaneous Exercise 4 | Q 4.2 | Page 114
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