I have the equations $$y=\frac{4\lambda}{\frac{1}{2}\lambda^2+2}\quad \text{and}\quad y=\frac{\lambda x}{-\frac{1}{2}\lambda ^2 + 2}$$ each representing a line. I'm asked to find the locus of the intersection between them.
I think I'm supposed to eliminate $\lambda$ between them but I don't know how to do it. I know from the book that the solution should be the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$, and that is confirmed by a geogebra plot.
Usually I just solve one of them for $\lambda$ and plug it into the other, but in this case both of them have $\lambda$ and $\lambda^2$ so I don't know how to solve it.
From the 2 equations we have: $x=\dfrac{4\cdot(4-\lambda^2)}{4+\lambda^2}$.
So, the coordinates of the points of intersection will be of the form: $$(x,y)=\left(\dfrac{4\cdot(4-\lambda^2)}{4+\lambda^2},\dfrac{8\lambda}{4+\lambda^2}\right).$$
$x^2=\dfrac{16\cdot(\lambda^4-8\lambda^2+16)}{16+8\lambda^2+\lambda^4}\implies \dfrac{x^2}{16}=\dfrac{\lambda^4-8\lambda^2+16}{16+8\lambda^2+\lambda^4}$
$y^2=\dfrac{64\lambda^2}{16+8\lambda^2+\lambda^4}\implies \dfrac{y^2}{4}=\dfrac{16\lambda^2}{16+8\lambda^2+\lambda^4}$
Finally, we take: $$\dfrac{x^2}{16}+\dfrac{y^2}{4}=\dfrac{\lambda^4-8\lambda^2+16}{16+8\lambda^2+\lambda^4}+\dfrac{16\lambda^2}{16+8\lambda^2+\lambda^4}=1$$