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# Find the equation of a curve passing through the point (0, –2) given that at any point  on the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.

Let x and y be the x-coordinate and y-coordinate of the curve respectively. We know that the slope of a tangent to the curve in the coordinate axis is given by the relation, According to the given information, we get: Integrating both sides, we get: Now, the curve passes through point (0, –2). ∴...

Let x and y be the x-coordinate and y-coordinate of the curve respectively.

We know that the slope of a tangent to the curve in the coordinate axis is given by the relation,

According to the given information, we get:

Integrating both sides, we get:

Now, the curve passes through point (0, –2).

∴ (–2)2 – 02 = 2C

⇒ 2C = 4

Substituting 2C = 4 in equation (1), we get:

y2x2 = 4

This is the required equation of the curve.

Expressing reality using equations gives birth to a mathematician

Given So Integrating both left and right sides of the above equation (0,-2) is on the curve So c=4 So curve is

Given $y×\frac{dy}{dx}=x$

So $ydy=xdx$

Integrating both left and right sides of the above equation

$y^{2}=x^{2}+&space;c$

(0,-2) is on the curve

So c=4

So curve is $y^{2}=x^{2}+4$

Related Questions

Find the general solution of the differential equation:

The given differential equation is: Now, integrating both sides, we get: This is the required general solution of the given differential equation.
Abishiek

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