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

प्रश्न

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"`.

बेरीज

उत्तर

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

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पाठ 6: Differential Equations - Exercise 6.5 [पृष्ठ २०७]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board

पाठ 6 Differential Equations
Exercise 6.5 | Q 3 | पृष्ठ २०७

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