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Question
If x = `(2bt)/(1 + t^2), y = a((1 - t^2)/(1 + t^2)), "show that" "dx"/"dy" = -(b^2y)/(a^2x)`.
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Solution
x = `(2bt)/(1 + t^2), y = a((1 - t^2)/(1 + t^2))`
Put t = tanθ.
Then x = `b((2tanθ)/(1 + tanθ)), y = a((1 - t^2)/(1 + tan^2θ))`
∴ x = b sin 2θ, y = a cos 2θ
∴ `x/b = sin2θ, y/a = cos2θ`
∴ `(x/b)^2 + (y/a)^2` = sin22θ + cos22θ
∴ `x^2/b^2 + y^2/a^2` = 1
Differentiating x and y w.r.t. y, we get
`(1)/b^2 xx 2x"dx"/"dy" + (1)/a^2 xx 2y` = 0
∴ `(2xdx)/(b^2dy) = (-2y)/a^2`
∴ `"dx"/"dy" = -(b^2y)/(a^2x)`
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