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प्रश्न
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.
उत्तर
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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