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प्रश्न
Find the equation of tangent and normal to the curve at the point on it.
y = x2 + 2ex + 2 at (0, 4)
उत्तर
y = x2 + 2ex + 2
∴ `dy/dx = d/dx(x^2 + 2e^x + 2)`
∴ `dy/dx = 2x + 2 × e^x + 0`
∴ `dy/dx = 2x + 2e^x`
∴ `(dy/dx)_("at" (0, 4))` = 2(0) + 2e0 = 2
⇒ slope of the tangent at (0, 4)
∴ the equation of the tangent at (0, 4) is
y – 4 = 2(x – 0)
∴ y – 4 = 2x
∴ 2x – y + 4 = 0
The slope of the normal at (0, 4)
= `(-1)/((dy/dx)_("at"(0, 4))) = -(1)/(2)`
∴ the equaion of the normal at (0, 4) is
y – 4 = `-(1)/(2)(x - 0)`
∴ 2y – 8 = – x
∴ x + 2y – 8 = 0
Hence, the equations of tangent and normal are 2x – y + 4 = 0 and x + 2y – 8 = 0 respectively.
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