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प्रश्न
Find the equations of tangents and normals to the following curves at the indicated points on them : 2xy + π sin y = `2pi "at" (1, pi/2)`
उत्तर
2xy + π sin y = 2π
Differentiating both sides w.r.t. x, we get
`2[xdy/dx + y.d/dx(x)] + pi cos ydy/dx` = 0
∴ `2xdy/dx + 2y xx 1 + pi cos ydy/dx` = 0
∴ `dy/dx = (-2y)/(2x + pi cos y)`
∴ `(dy/dx)_("at"(1, pi/2)) = (-2(pi/2))/(2(1) + pi cos pi/2)`
= `(-pi)/(2 + pi(0)`
= `-pi/(2)`
= slope of the tangent at `(1, pi/2)`
∴ the equation of the tangent at `(1, pi/2)` is
`y - pi/(2) = - pi/(2)(x - 1)`
∴ 2y – π = – πx + π
∴ πx + 2y – 2π = 0
The slope of normal at `(1, pi/2)`
= `(-1)/((dy/dx)_("at" (1, pi/2))`
= `(-1)/((- pi/2)`
= `(2)/pi`
∴ the equation of the normal at `(1, pi/2)` is
`y - pi/(2) = (2)/pi(x - 1)`
∴ `piy - pi^2/(2)` = 2x – 2
∴ `2piy - pi^2` = 4x – 4
∴ 4x – 2πy + π2 – 4 = 0
Hence, the equations of tangent and normal are
πx + 2y – 2π = 0 and 4x – 2πy + π2 – 4 = 0 respectively.
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