Maharashtra State BoardHSC Arts 12th Board Exam
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Find the equation of the curve passing through the point (32,2) having a slope of the tangent to the curve at any point (x, y) is -4x9y4x9y. - Mathematics and Statistics

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Sum

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`

∴ 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.

Concept: Methods of Solving First Order, First Degree Differential Equations - Linear Differential Equations
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