Question

Find tthe equation of the curve for which product of slope of tangent with its $y$-co-ordinate at any point $(x,y)$ to the curve is $x$-co-ordinate and the curve is passed through $\left(2.1\right)$.

Answer 1

Given that the product of $y$- coordinate of point $(x,y)$ and slope of tangent is its $x$-coordinate.

Therefore,

$y.\frac{dy}{dx}=x$

$\Rightarrow ydy=xdx$

Integrating both sides, we have

$\int ydy=\int xdx$

$\Rightarrow \frac{y^2}{2}+C_1=\frac{x^2}{2}+C_2$

$\Rightarrow x^2-y^2=2(C_1-C_2)$

$\Rightarrow x^2+y^2=C.....\left(1\right)[\because 2(C_1-C_2)=\text{Constant}]$

Given that the curve passes through the point $(2,1)$.

Therefore, the point $(2,1)$ will satisfy the equation of curve.

Therefore,

${\left(2\right)}^2+{\left(1\right)}^2=C$

$\Rightarrow C=5$

Substituting the value of $C$ in ${eq}^n\left(1\right)$, we have

$x^2+y^2=5$

Therefore, the equation of the curve is $x^2+y^2=5$.