Question

If ax² + 2hxy + by² + 2gx + 2fy + c = 0, then show that `"dy"/"dx" * "dx"/"dy"` = 1 

Answer 1

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.