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Question
If x = `asqrt(secθ - tanθ), y = asqrt(secθ + tanθ), "then show that" "dy"/"dx" = -y/x`.
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Solution
x = `asqrt(secθ - tanθ), y = asqrt(secθ + tanθ)`
∴ `x/a = sqrt(secθ - tanθ), y/a = sqrt(secθ + tanθ)`
∴ `sec θ - tanθ = x^2/a^2` ...(1)
`sec θ + tanθ = y^2/a^2` ...(2)
Adding (1) and (2), we get
2secθ = `x^2/a^2 + y^2/a^2`
= `(x^2 + y^2)/a^2`
∴ secθ = `(x^2 + y^2)/(2a^2)`
Subtracting (1) from (2), we get
2tanθ = `y^2/a^2 - x^2/a^2`
= `(y^2 - x^2)/a^2`
∴ tanθ = `(y^2 - x^2)/(2a^2)`
∴ sec2θ - tan2θ = 1 gives,
`((x^2 + y^2)/(2a^2))^2 - ((y^2 - x^2)/(2a^2))^2` = 1
∴ (x2 + y2)2 - (y2 - x2)2 = 4a4
∴ (x4 + 2x2y2 + y4) - (y4 - 2x2y2 + x4) = 4a4
∴ 4x2y2 = 4a4
∴ x2y2 = a4
Differentiating both sides w.r.t. x, we get
`x^2."d"/"dx"(y^2) + y^2."d"/"dx"(x^2)` = 0
∴ `x^2 xx 2y"dy"/"dx" + y^2 xx 2x` = 0
∴ `2x^2y"dy"/"dx"` = -2xy2
∴ `"dy"/"dx" = -y/x`.
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