Question

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

xy = c² at `("ct", "c"/"t")` where t is parameter.

Answer 1

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