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Question
If ax2 + 2hxy + by2 = 0, then show that `("d"^2"y")/"dx"^2` = 0
Solution
ax2 + 2hxy + by2 = 0 ....(i)
Differentiating both sides w.r.t. x, we get
`"a"("2x") + "2h" * "d"/"dx" ("xy") + "b"("2y") "dy"/"dx" = 0`
∴ `2"ax" + 2"h" ["x" * "dy"/"dx" + "y"(1)] + 2"by" "dy"/"dx" = 0`
∴ `2"ax" + 2"hx" "dy"/"dx" + 2"hy" + 2"by" "dy"/"dx" = 0`
∴ `2 "dy"/"dx" ("hx" + "by") = - 2"ax" - 2"hy"`
∴ `2 "dy"/"dx" = (-2("ax" + "hy"))/("hx" + "by")`
∴ `"dy"/"dx" = (- ("ax" + "hy"))/("hx" + "by")` ....(i)
ax2 + 2hxy + by2 = 0
∴ ax2 + hxy + hxy + by2 = 0
∴ x(ax + hy) + y(hx + by) = 0
∴ y(hx + by) = - x(ax + hy)
∴ `"y"/"x" = (- ("ax" + "hy"))/("hx" + "by")` ...(ii)
From (i) and (ii), we get
`"dy"/"dx" = "y"/"x"` ....(iii)
Again, differentiating both sides w.r.t. x, we get
`("d"^2"y")/"dx"^2 = ("x" * "dy"/"dx" - "y" * "d"/"dx" ("x"))/"x"^2`
`= ("x" * ("y"/"x") - "y"(1))/"x"^2` ....[From (iii)]
`= ("y - y")/"x"^2`
`= 0/"x"^2`
∴ `("d"^2"y")/"dx"^2 = 0`
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