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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"`.

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#### 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`

∴ 9y^{2} = - 4x^{2} + 18c_{1}

∴ 4x^{2} + 9y^{2} = c_{1} where c = 18c_{1} .....(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 4x^{2} + 9y^{2} = 36.

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