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Find the equation of tangent and normal to the following curve. xy = c2 at (ct,ct) where t is parameter. - Mathematics and Statistics

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Sum

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

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

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Solution

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

[Note: Answer in the textbook is incorrect.]

Concept: Introduction of Derivatives
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 4 Applications of Derivatives
Miscellaneous Exercise 4 | Q 4.1 | Page 114
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