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Find the equation of tangent and normal to the curve y = x2 + 5 where the tangent is parallel to the line 4x – y + 1 = 0. - Mathematics and Statistics

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Question

Find the equation of tangent and normal to the curve y = x2 + 5 where the tangent is parallel to the line 4x – y + 1 = 0.

Sum

Solution

Le P(x1, y1) be the point on the curve y = x2 + 5 where the tangent is parallel to the line 4x – y + 1 = 0.

Differentiating w.r.t.x we get

`dy/dx = d/dx(x^2 + 5)` = 2x + 0 = 2x

∴ `(dy/dx)_(at(x_1.y_1)` = 2x1

Slope of the tangent at (x1, y1)

Let m1 = 2x1,

The slope of line 4x – y + 1 = 0 is

m2 = `(-4)/(-1)` = 4

∵ The tangent at P(x1, y1) is parallel to the line

4x – y + 1 = 0, m1 = m2

∴ 2x1 = 4

∴ x1 = 2

Since (x1, y1) lies on the curve y = x2 + 5, y1 = x12 + 5

∴ y1 = (2)2 + 5 = 9  ...[∵ x1 = 2]

∴ The coordinates of the point are (2, 9) and the slope of the tangent= m1 = m2 = 4

∴ Equation of the tangent is y – 9 = 4(x – 2)

∴ y – 9 = 4(x – 2)

∴ y – 9 = 4x – 8

∴ 4x – y + 1 = 0

Slope of the normal = `(-1)/m_1 = -1/4`

∴ Equation of normal at (2, 9) is y – 9 = `(-1)/4(x - 2)`

∴ 4y – 36 = –x + 2

∴ x + 4y – 38 = 0

Hence, the equation of tangent and normal are 4x – y + 1 = 0 and x + 4y – 38 = 0 respectively.

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