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Question
If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, then show that `"dy"/"dx" * "dx"/"dy"` = 1
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Solution
Given that: ax2 + 2hxy + by2 + 2gx + 2fy + c = 0.
Differentiating both sides w.r.t. x
`"d"/"dx" ("a"x^2 + 2"h"xy + "b"y^2 + 2"g"x + 2"f"y + "c") = "d"/"dx" (0)`
⇒ `"a"*2x + 2"h"(x * "dy"/"dx" + y*1) + "b"*2*y* "dy"/"dx" + 2"g"*1 + 2"f"* "dy"/"dx" + 0` = 0
⇒ `2"a"x + 2"h"x * "dy"/"dx" + 2"h"y + 2"b"y * "dy"/"dx" + 2"g" + 2"f" * "dy"/"dx"` = 0
⇒ `2"h"x * "dy"/"dx" + 2"b"y "dy"/"dx" + 2"f" "dy"/"dx"` = – 2ax – 2hy – 2g
⇒ `(2"h"x + 2"b"y + 2"f") "dy"/"dx"` = – 2(ax + hy + g)
⇒ `2("h"x + "b"y + "f") "dy"/"dx"` = = – 2(ax + hy + g)
⇒ `"dy"/"dx" = (-2("a"x + "h"y + "g"))/(2("h"x + "b"y + "f"))`
⇒ `"dy"/"dx" = (-("a"x + "h"y + "g"))/(("h"x + "b"y + "f"))`
Now, differentiating the given equation w.r.t. y.
`"d"/"dy" ("a"x^2 + 2"h"xy + "b"y^2 + 2"g"x + 2"f"y + "c") = "d"/"dy"(0)`
⇒ `2"a"x* "dx"/"dy" + 2"h" (y * "dx"/"dy" + x*1) + 2"b"y + 2"g" * "dx"/"dy" + 2"f" * 1 + 0` = 0
⇒ `2"a"x * "dx"/"dy" + 2"h"y * "dx"/"dy" + 2"h"x + 2"b"y + 2"g" * "dx"/"dy" + 2"f"` = 0
⇒ `2"a"x "dx"/"dy" + 2"h"y * "dx"/"dy" + 2"g" * "dx"/"dy"` = – 2hx – 2by – 2f
⇒ `(2"a"x + 2"h"y + 2"g") "dx"/"dy"` = = – 2hx – 2by – 2f
⇒ `"dx"/"dy" = (-2"h"x - 2"b"y - 2"f")/(2"a"x + 2"h"y + 2"g")`
⇒ `"dx"/"dy" = (-2("h"x + "b"y + "f"))/(2("a"x + "h"y + "g"))`
⇒ `"dx"/"dy" = (-("h"x + "b"y + "f"))/(("a"x + "h"y + "g"))`
∴ `"dy"/"dx" * "dx"/"dy" = [(-("a"x + "h"y + "g"))/(("h"x + "b"y + "f"))][(-("h"x + "b"y + "f"))/(("a"x + "h"y + "g"))]` = 1
Hence, `"dy"/"dx" * "dx"/"dy"` = 1.
Hence proved.
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