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प्रश्न
Find the equation of tangent and normal to the following curve.
x = `1/"t", "y" = "t" - 1/"t"`, at t = 2
उत्तर
x = `1/"t", "y" = "t" - 1/"t"`
∴ `"dx"/"dt" = -1/"t"^2, "dy"/"dt" = 1 + 1/"t"^2`
∴ `"dy"/"dx" = ("dy"/"dt")/("dx"/"dt") = (1 + 1/"t"^2)/(-1/"t"^2) = - "t"^2 - 1`
Slope of tangent at t = 2 is
`("dy"/"dx")_("t" = 2)` = - (2)2 - 1 = - 5
∴ Point is (x1, y1) = `(1/2, 2 - 1/2) = (1/2, 3/2)`
Equation of tangent at `(1/2, 3/2)`
`"y" - 3/2 = - 5("x" - 1/2)`
∴ 2y - 3 = - 5(2x - 1)
∴ 10x + 2y = 8
∴ 5x + y = 4
∴ 5x + y - 4 = 0
Slope of normal at t = 2 is `(-1)/("dy"/"dx")_("t" = 2)` = `(-1)/-5 = 1/5`
Equation of normal at `(1/2, 3/2)` is
`"y" - 3/2 = 1/5("x" - 1/2)`
∴ `("2y" - 3)/2 = ("2x" - 1)/10`
∴ 10y - 15 = 2x - 1
∴ 2x - 10y + 14 = 0
∴ x - 5y + 7 = 0
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