Question

Let $F$ be the family of curves given by

$x^2+2hxy+y^2=1,-1<h<1.$

Then, the differential equation for the family of orthogonal trajectories to $F$ is

1. $\left(x^{2}y-y^3+y\right)\frac{dy}{dx}-\left(x{y}^2-x^3+x\right)=0$
2. $\left(x^{2}y-y^3+y\right)\frac{dy}{dx}+\left(x{y}^2-x^3+x\right)=0$
3. $\left(x^{2}y+y^3+y\right)\frac{dy}{dx}-\left(x{y}^2+x^3+x\right)=0$
4. $\left(x^{2}y+y^3+y\right)\frac{dy}{dx}+\left(x{y}^2+x^3+x\right)=0$

Answer: $\left(x^{2}y-y^3+y\right)\frac{dy}{dx}-\left(x{y}^2-x^3+x\right)=0$

Solution :-

We have, given faimly of curves

Differentiating eq(1) w. r. to $x$ , we get

Now, from eq(1), we have

Now, put the value of $h$ in eq(2), we have

Now, replace $\frac{dy}{dx}$ by $-\frac{dx}{dy}$ in eq(3), we get

So, correct answer is option-1, $\to \left(x^{2}y-y^3+y\right)\frac{dy}{dx}-\left(x{y}^2-x^3+x\right)=0$