# 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

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".

#### 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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