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Question
Find the equation of the curve passing through the point `(3/sqrt2, sqrt2)` having a slope of the tangent to the curve at any point (x, y) is -`"4x"/"9y"`.
Solution
Let A(x, y) be the point on the curve y = f(x).
Then the slope of the tangent to the curve at point A is `"dy"/"dx"`.
According to the given condition
`"dy"/"dx" = - "4x"/"9y"`
∴ y dy = `- 4/9 "x dx"`
Integrating both sides, we get
`int "y dy" = - 4/9 int "x dx"`
∴ `"y"^2/2 = - 4/9 * "x"^2/2 + "c"_1`
∴ 9y2 = - 4x2 + 18c1
∴ 4x2 + 9y2 = c1 where c = 18c1 .....(1)
This is the general equation of the curve.
But the required curve is passing through the point `(3/sqrt2, sqrt2)`.
∴ by putting x = `3/sqrt2` and y = `sqrt2` in (1), we get
`4(3/sqrt2)^2 + 9(sqrt2)^2 = "c"`
∴ 18 + 18 = c
∴ c = 36
∴ from (1), the equation of the required curve is 4x2 + 9y2 = 36.
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