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प्रश्न
Find the point on the curve y = `sqrt(x - 3)` where the tangent is perpendicular to the line 6x + 3y – 5 = 0.
उत्तर
Let the required point on the curve y = `sqrt(x - 3)` be P(x1, y1).
Differentiating y = `sqrt(x - 3)` w.r.t.x., we get
`(dy)/(dx) = d/(dx)(sqrt(x - 3))`
= `1/(2sqrt(x - 3)).d/(dx)(x - 3)`
= `(1 xx (1 - 0))/(2sqrt(x - 3))`
= `1/(2sqrt(x - 3))`
∴ Slope of the tangent at (x1, y1)
= `(dy/dx)_(at(x_1, y_1))`
= `1/(2sqrt(x_1 - 3))`
Since, this tangent is perpendicular to 6x + 3y – 5 = 0 whose slope is `(-6)/3` = –2.
Slope of the tangent = `(-1)/(-2) = 1/2`
∴ `1/(2sqrt(x_1 - 3)) = 1/2`
∴ `sqrt(x_1 - 3)` = 1
∴ x1 – 3 = 1
∴ x1 = 4
Since, (x1, y1) lies on y = `sqrt(x - 3)`, y1 = `sqrt(x_1 - 3)`
When x1 = 4, y1 = `sqrt(4 - 3)` = ±1
Hence, the required points are (4, 1) and (4, –1).
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