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प्रश्न
Show that `"dy"/"dx" = y/x` in the following, where a and p are constants : `tan^-1((3x^2 - 4y^2)/(3x^2 + 4y^2))` = a2
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उत्तर
`tan^-1((3x^2 - 4y^2)/(3x^2 + 4y^2))` = a2
∴ `(3x^2 - 4y^2)/(3x^2 + 4y^2)` = tana2 = k ...(Say)
∴ 3x2 – 4y2 = 3kx2 + 4ky2
∴ (4k + 4)y2 = (3 – 3k)x2
∴ `y^2/x^2 = (3 - 3k)/(4k + 4)`
∴ `y/x = sqrt((3 - 3k)/(4k + 4)`, a constant
Differentiating both sides w.r.t. x, we get
`"d"/"dx"(y/x)` = 0
∴ `(x"dy"/"dx" - y."d"/"dx"(x))/x^2` = 0
∴ `x"dy"/"dx" - y xx 1` = 0
∴ `x."dy"/"dx"` = y
∴ `"dy"/"dx" = y/x`.
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