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प्रश्न
Find the equations of tangent and normal to the curve y = x2 + 5 where the tangent is parallel to the line 4x − y + 1 = 0.
उत्तर
Equation of the curve is y = x2 + 5
Differentiating w.r.t. x, we get
`"dy"/"dx"` = 2x
Slope of the tangent at P(x1, y1) is
`("dy"/"dx")_(("x"_1,"x"_2)` = 2x1
According to the given condition, the tangent is parallel to 4x – y + 1 = 0
Now, slope of the line 4x – y + 1 = 0 is 4.
∴ Slope of the tangent = `"dy"/"dx" = 4`
∴ 2x1 = 4
∴ x1 = 2
P(x1, y1) lies on the curve y = x2 + 5
∴ y1 = (2)2 + 5
∴ y1 = 9
∴ The point on the curve is (2, 9).
∴ Equation of the tangent at (2, 9) is
∴ (y – 9) = 4(x – 2)
∴ y – 9 = 4x – 8
∴ 4x - y + 1 = 0
Slope of the normal at (2, 9) is `1/("dy"/"dx")_((2,9)` = `(- 1)/4`
∴ Equation of the normal of (2, 9) is
(y - 9) = `(-1)/4`(x - 2)
∴ 4y - 36 = - x + 2
∴ x + 4y - 38 = 0
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