Hello, bherila!
Find the equation of the curve that goes through (1,3)
for which any tangent and the line from the origin to the point of contact
make with the ordinate a triangle having area $\displaystyle A$. Code:

* * y = f(x)
 *
 * *
 * P(p,q)
 o /
 o /
 o /
B o /
 /
 /
 /
  *           
O

We have a function: $\displaystyle y \,=\,f(x)$
Let $\displaystyle P(p,q)$ be a point on $\displaystyle y \,=\,f(x).$
$\displaystyle OP$ is the segment from the origin to point $\displaystyle P.$
$\displaystyle BP$ is the tangent at $\displaystyle P$.
. . The slope of this tangent is $\displaystyle y'$ and it contains $\displaystyle P(p,q).$
. . Its equation is: .$\displaystyle y  q \:=\:y'(x  p)\quad\Rightarrow\quad y \:=\:y'x  y'p + q$
. . Its yintercept is: .$\displaystyle B(0,\,q  py')$
The base of $\displaystyle \Delta OBP$ is: .$\displaystyle OB \:=\:q  py'$
The height of the triangle is: .$\displaystyle p$
The area of $\displaystyle \Delta OBP\!:\;\;\frac{1}{2}(q  py')p \;=\;A$
\. . which simplifies to: .$\displaystyle p^2y'  pq \:=\:2A$
Divide by $\displaystyle p^2\!:\;\;y'  \frac{1}{p}\,q \:=\:\frac{2A}{p^2}$
Hence, we have: .$\displaystyle \frac{dy}{dx}  \frac{1}{x}\,y \;=\;\frac{2A}{x^2}$
Integrating factor: .$\displaystyle I \:=\:e^{\int\left(\frac{1}{x}\right)dx} \:=\:e^{\ln x} \:=\:e^{\ln\frac{1}{x}} \:=\:\frac{1}{x}$
So we have: .$\displaystyle \frac{1}{x}\,\frac{dy}{dx}  \frac{1}{x^2}\,y \;=\;\frac{2A}{x^3} $
Then: .$\displaystyle \frac{d}{dx}\left(\frac{1}{x}\!\cdot\!y\right) \;=\;2Ax^{3}$
Integrate: .$\displaystyle \frac{1}{x}\!\cdot\!y \;=\;Ax^{2} + C$
Multiply by $\displaystyle x\!:\;\;y \;=\;\frac{A}{x} + Cx$
Since $\displaystyle (1,3)$ is on the curve,
. . $\displaystyle 3 \;=\;\frac{A}{1} + C\quad\Rightarrow\quad C \:=\:A + 3$
Therefore: .$\displaystyle \boxed{y \;=\;\frac{A}{x} + (A + 3)x}$