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Sum

**Find the equation of tangent and normal to the following curve.**

xy = c^{2} at `("ct", "c"/"t")` where t is parameter.

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#### Solution

Equation of the curve is xy = c^{2}

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")`

∴ yt^{2} - ct = - x + ct

∴ x + yt^{2} - 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 = xt^{3} - ct^{4}

∴ t^{3}x - yt - (t^{4} - 1)c = 0

#### Notes

The answer in the textbook is incorrect.

Concept: Introduction of Derivatives

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