Question

If x = a sec θ, y = b tan θ, then `("d"^2"y")/("dx"^2)` at θ = `π/6` is:

विकल्प

• `(-3sqrt3"b")/"a"^2`

• `(-2sqrt3"b")/"a"`

• `(-3sqrt3"b")/"a"`

• `(-"b")/(3sqrt3"a"^2)`

Answer 1

`(-3sqrt3"b")/"a"^2`

Explanation:

x = a sec θ ⇒ `"dx"/("d"θ)` = a sec θ tan θ

y = b tan θ ⇒ `"dy"/("d"θ)` = b sec² θ

∴ `"dy"/"dx" = "b"/"a" "cosec θ"`

⇒ `("d"^2"y")/("dx"^2) = (-"b")/"a" "cosec θ". cot θ. ("d"θ)/"dx" = (-"b")/"a"^2 cot^3 θ`

∴ `[("d"^2"y")/("dx"^2)]_(θ = π/6) = (-3sqrt3"b")/"a"^2`