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प्रश्न
If x = `(4"t")/(1 + "t"^2)`, y = `3((1 - "t"^2)/(1 + "t"^2))`, then show that `("d"y)/("d"x) = (-9x)/(4y)`
उत्तर
x = `(4"t")/(1 + "t"^2)`
Differentiating both sides w.r.t. ‘t’, we get
`("d"x)/"dt" = "d"/"dt" ((4"t")/(1 + "t"^2))`
= `((1 + "t"^2)*"d"/"dt"(4"t") - 4"t"*"d"/"dt"(1 + "t"^2))/(1 + "t"^2)^2`
= `((1 + "t"^2)(4) - 4"t"(0 + 2"t"))/(1 + "t"^2)^2`
= `(4 + 4"t"^2 - 8"t"^2)/(1 + "t"^2)^2`
= `(4 - 4"t"^2)/(1 + "t"^2)^2`
= `(4(1 - "t"^2))/(1 + "t"^2)^2`
y = `3((1 - "t"^2)/(1 + "t"^2))`
`("d"y)/"dt" = 3*"d"/"dt"((1 - "t"^2)/(1 + "t"^2))`
= `3[((1 + "t"^2)*"d"/"dt"(1 - "t"^2) - (1 - "t"^2)*"d"/"dt"(1 + "t"^2))/(1 + "t"^2)]`
= `3[((1 + "t"^2)(0 - 2"t") - (1 - "t"^2)(0 + 2"t"))/(1 + "t"^2)^2]`
= `3[(-2"t"(1 + "t"^2) - 2"t"(1 - "t"^2))/(1 + "t"^2)^2]`
= `3(- 2"t")[(1 + "t"^2 + 1 - "t"^2)/(1 + "t"^2)^2]`
= `- 6"t" xx 2/(1 + "t"^2)^2`
= `(-12"t")/(1 + "t"^2)^2`
∴ `("d"y)/("d"x) = (("d"y)/("dt"))/(("d"x)/("dt"))`
= `((-12"t")/((1 + "t"^2)^2))/((4(1 - "t"^2))/((1 + "t"^2)^2)`
= `("d"y)/("d"x) = (-3"t")/(1 - "t"^2)` ......(i)
Also, `(-9x)/(4y) = (-9((4"t")/(1 + "t"^2)))/(4 xx 3((1 - "t"^2)/(1 + "t"^2))`
= `(-3"t")/(1 -"t"^2)` ......(ii)
From (i) and (ii), we get
`("d"y)/("d"x) = (-9x)/(4y)`
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