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प्रश्न
Solve the following : Find the equation of the tangent and normal drawn to the curve y4 – 4x4 – 6xy = 0 at the point M (1, 2).
उत्तर
y4 – 4x4 – 6xy = 0
Differentiating both sides w.r.t x, we get
`4y^3dy/d - 4 xx 4x^3 - 6[xdy/dx + y.d/dx(x)]` = 0
∴ `4y^3dy/dx - 16x^3 - 6xdy/dx - 6y xx 1` = 0
∴ `(4y^3 - 6x)dy/dx` = 16x3 + 6y
∴ `dy/dx = (16x^3 + 6y)/(4y^3 - 6x)`
= `(8x^3 + 3y)/(2y^3 - 3x)`
∴ `(dy/dx)_("at" (1, 2)) = (8(1)^3 + 3(2))/(2(2)^3 - 3(1))`
= `(8 + 6)/(16 - 3)`
= `(14)/(13)`
= slope of the tangent at (1, 2)
∴ the equation of the tangent at M (1, 2) is
y – 2 = `(14)/(13)(x - 1)`
∴ 13y – 26 = 14x – 14
∴ 14x – 13y + 12 = 0
The slope of normal at (1, 2)
= `(-1)/((dy/dx)_("at" (1, 2)`
= `(-1)/((14/13)`
= `-(13)/(14)`
∴ the equation of normal at M (1, 2) is
y – 2 = `-(13)/(14)(x - 1)`
∴ 14y – 28 = –13x + 13
∴ 13x + 14y – 41 = 0
Hence, the equations of tangent and normal are
14x – 13y + 12 = 0 and 13x + 14y – 41 = 0 respecttively.
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