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प्रश्न
Find the equation of tangent and normal to the following curve.
xy = c2 at `("ct", "c"/"t")` where t is parameter.
उत्तर
Equation of the curve is xy = c2
Differentiating w.r.t. x, we get
`"x" "dy"/"dx" + "y" = 0`
∴ `"dy"/"dx" = (- "y")/"x"`
∴ slope of tangent at `("ct", "c"/"t")` is
`("dy"/"dx")_(("ct", "c"/"t")` = `((- "c")/"t")/"ct" = (- 1)/"t"^2`
Equation of tangent at `("ct", "c"/"t")` is
`("y" - "c"/"t") = (-1)/"t"^2 ("x" - "ct")`
∴ yt2 - ct = - x + ct
∴ x + yt2 - 2ct = 0
Slope of normal =`(-1)/((-1)/"t"^2) = "t"^2`
Equation of normal at `("ct", "c"/"t")` is
`("y" - "c"/"t") = "t"^2 ("x" - "ct")`
∴ yt - c = xt3 - ct4
∴ t3x - yt - (t4 - 1)c = 0
Notes
The answer in the textbook is incorrect.
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