The chord of contact of the point $$(x_1,y_1)$$ with respect to the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ cuts the axes at $L$ and $M$. If the locus of the mid-point of $LM$ is the circle $$x^2+y^2=1$$, find the locus of $(x_1,y_1)$
MY APPROACH
Equation of chord of contact is $$b^2xx_1+a^2yy_1=a^2b^2\tag{1}$$ $\implies$ $\frac{x}{\left(\frac{a^2}{x_1}\right)}+\frac{y}{\left(\frac{b^2}{y_1}\right)}=1$
Points where $(1)$ cuts the x-axes are $$L\left(\frac{a^2}{x_1},0\right)$$ $$M\left(0,\frac{b^2}{y_1}\right)$$ Mid-point $G(h,k)$ of chord $LM$
$$G(h,k)=\left(\frac{a^2}{2x_1},\frac{b^2}{2y_1}\right)\tag{2}$$
Locus of $(2)$ is given as $$x^2+y^2=1\tag{3}$$
Then,$(2)$ must satisfy $(3)$ $$\implies\left(\frac{a^2}{2x_1}\right)^2+\left(\frac{b^2}{2y_1}\right)^2=1$$ $$\implies a^4y_1^2+b^4x_1^2=4x_1^2y_1^2$$
In general $x_1=x$ and $y_1=y$ $$\therefore a^4y^2+b^4x^2=4x^2y^2$$ is the required locus.
[1] My doubt starts from $(3)$ to the final step
Is it OK how I found the locus by generalizing $x_1$ and $y_1$ as $x$ and $y$ instead of eliminating them via a certain substitution?
[2] If [1] is OK and my whole approach is correct. Could I have reached to the solution by a different approach?.
I appreciate your attention and contribution