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प्रश्न
Find the equations of tangents and normals to the following curves at the indicated points on them : x3 + y3 – 9xy = 0 at (2, 4)
उत्तर
x3 + y3 – 9xy = 0
Differentiating both sides w.r.t. x, we get
`3x^2 + 3y^2dy/dx - 9[xdy/dx + y.d/dx(x)]` = 0
∴ `3x^2 + 3y^2dy/dx - 9xdy/dx - 9y` = 0
∴ `(3y^2 - 9x)dy/dx` = 9y – 3x2
∴ `dy/dx = (9y - 3x^2)/(3y^2 -9x)`
∴ `(dy/dx)_("at"(2, 4)) = (9(4) - 3(2)^2)/(3(4)^2 - 9(2)`
= `(36 - 12)/(48 - 18)`
= `(24)/(30)`
= `(4)/(5)`
= Slope of the tangent at (2, 4)
∴ The equation of the tangent at (2, 4) is `y - 4 = (4)/(5)(x - 2)`
∴ 5y – 20 = 4x – 8
∴ 4x – 5y + 12 = 0
The slope of normal at (2, 4)
= `(-1)/((dy/dx)_("at"(2, 4)`
= `-(5)/(4)`
∴ The equation of the normal at (2, 4) is
`y - 4 = -(5)/(4)(x - 2)`
∴ 4y – 16 = – 5x + 10
∴ 5x + 4y – 26 = 0
Hence, the equation of tangent and normal are
4x – 5y + 12 = 0 and 5x + 4y – 26 = 0 respectively.
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