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प्रश्न
Find the equations of tangents and normals to the following curves at the indicated points on them: `x^2 - sqrt(3)xy + 2y^2 = 5 "at" (sqrt(3),2)`
उत्तर
`x^2 - sqrt(3)xy + 2y^2` = 5
Differentiating both sides w.r.t. x, we get
`2x - sqrt(3)[xdy/dx + y.d/dx(x)] + 2 xx 2ydy/dx` = 0
∴ `2x - sqrt(3)xdy/dx - sqrt(3)y xx 1 + 4ydy/dx` = 0
∴ `(4y - sqrt(3)x)dy/dx = sqrt(3)y - 2x`
∴ `dy/dx = (sqrt(3)y - 2x)/(4y - sqrt(3)x)`
= `(2x - sqrt(3)y)/(sqrt(3)x - 4y)`
∴ `(dy/dx)_("at"(sqrt(3), 2)) = (2sqrt(3) - sqrt(3)(2))/(sqrt(3)(sqrt(3)) - 4(2)` = 0
= slope of the tangent at `(sqrt(3), 2)`
∴ the equation of the tangent at `(sqrt(3), 2)` is
y – 2 = `0(x - sqrt(3))`
∴ y – 2 = 0
∴ y = 2
The slope of normal at `(sqrt(3), 2)`
= `(-1)/((dy/dx)_("at"(sqrt(3), 2))), "where" (dy/dx)_("at"(sqrt(3),2))` = 0
∴ the slope of normal at `(sqrt(3), 2)` does not exist.
∴ normal is parallel to Y-axis.
∴ equation of the normal is of the form x = k
Since, it pases through the point `(sqrt(3), 2), k = sqrt(3)`
∴ equation of the normal is x = `sqrt(3)`.
Hence, the equations of tangent and normal are
y = 2 and x = `sqrt(3)` respectively.
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