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प्रश्न
x = `"e"^theta (theta + 1/theta)`, y= `"e"^-theta (theta - 1/theta)`
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उत्तर
Given that, x = `"e"^theta (theta + 1/theta)`, y= `"e"^-theta (theta - 1/theta)`
Differentiating both the parametric functions w.r.t. θ
`"dx"/("d"theta) = "e"^theta(1 - 1/theta^2) + (theta + 1/theta)*"e"^theta`
`"dx"/("d"theta) = "e"^theta (1 - 1/theta^2 + theta + 1/theta)`
⇒ `"e"^theta ((theta^2 - 1 + theta^3 + theta)/theta^2)`
= `("e"^theta(theta^3 + theta^2 + theta - 1))/theta^2`
y = `"e"^-theta(theta - 1/theta)`
`"dy"/("d"theta) = "e"^-theta(1 + 1/theta^2) + (theta - 1/theta) * (-"e"^-theta)`
`"dy"/("d"theta) = "e"^-theta (1 + 1/theta^2 - theta + 1/theta)`
⇒ `"e"^-theta ((theta^2 + 1 - theta^3 + theta)/theta^2)`
= `"e"^-theta ((-theta^3 + theta^2 + theta + 1))/theta^2`
∴ `"dy"/"dx" = (("dy")/("d"theta))/(("d"x)/("d"theta))`
= `("e"^-theta ((-theta^3 + theta^2 + theta + 1)/theta^2))/("e"^theta ((theta^3 + theta^2 + theta + 1)/theta^2))`
= `"e"^(-2theta) ((-theta^3 + theta^2 + theta + 1)/(theta^3 + theta^2 + theta - 1))`
Hence, `"dy"/"dx" = "e"^(-2theta) ((-theta^3 + theta^2 + theta + 1)/(theta^3 + theta^2 + theta - 1))`.
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