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