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Question
Find the tangent and normal to the following curves at the given points on the curve
y = x4 + 2ex at (0, 2)
Solution
y = x4 + 2ex at (0, 2)
Differentiating w.r.t. ‘x’
`("d"y)/("d"x)` = 4x3 + 2ex
Slope of the tangent ‘m’
`(("d"y)/("d"x))_(((0, 2))` = 4(0)3 + 2e0 = 2
Slope of the Normal `- 1/"m" = - 1/2`
Equation of tangent is
y – y1 = m(x – x1)
⇒ y – 2 = 2(x – 0)
⇒ y – 2 = 2x
⇒ 2x – y + 2 = 0
Equation of Normal is
y – y1 = `- 1/"m"` (x – x1)
y – 2 = `- 1/2` (x – 0)
2y – 4 = – x
x + 2y – 4 = 0
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