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$y = \frac{\lambda x}{2}$

$x+2 = 2 \lambda y$

$\frac{x^2}{4}+y^2=1$

I attempted to compute $y^2=\frac{\lambda^2x^2}{4}$ and $x^2= (2 \lambda y-2)( 2 \lambda y-2)=4\lambda^2y^2-8\lambda y+4$ and I was hoping to plug into $\frac{x^2}{4}+y^2=1$, but it will still be in terms of $\lambda$ and $y$, so I can't exactly solve for either one of them.

Would greatly appreciate any advice from the community.

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Substitute $y = \frac{\lambda x}{2}$ (eqn 1) in $x + 2 = 2 \lambda y$ (eqn 2) to obtain $$x = \frac{2}{\lambda^{2}-1} (\text{eqn } 1').$$ Solve for $y$ from eqn 1 to obtain $$y = \frac{\lambda}{\lambda^{2}-1} (\text{eqn } 2').$$

Plug the values of $x$ and $y$ in eqn 3: $\frac{x^{2}}{4} + y^{2} = 1$ and solve for $\lambda$ to obtain the values $$\lambda = -\sqrt{3}, \sqrt{3}.$$

Plug the values of $\lambda$ in eqn 1' and eqn 2' to get the solutions $$\left ( 1,\sqrt{\frac{3}{2}} \right ), \left ( 1,-\sqrt{\frac{3}{2}} \right ). $$

You can also check the solutions by plugging the values of $x$ and $y$ in eqn 3 equaling to $1$ or not, that is, the right-hand side.

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  • $\begingroup$ You definitely got the correct answer, but I'm not able to follow how you solved for $y=\frac{\lambda}{\lambda^2-1}$. How did you get that? $\endgroup$ Oct 12 '16 at 3:45
  • $\begingroup$ Substitute away $x$ in eqn 1 by putting in the value of $x$ from eqn 1'. $\endgroup$
    – OGC
    Oct 12 '16 at 11:16
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If $\lambda = 0$, $y = 0, \ x=-2$ and the third equation holds. If not, then $x = \frac{2y}{\lambda}$ and from the third equation you get now a polynomial for y. Solve it and the plug everything in the second equation. You should now have a polynomial in $\lambda$.

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Using $y=\frac{\lambda x }{2}$ in the other two equations gives $$\frac{x^2}{4} + \frac{\lambda^2 x^2}{4} =1 \text{ and } x+2 = 2 \lambda^2 \frac{x}{2}$$ These equations become $$(1+\lambda^2) x^2 -4 = 0 \text{ and } x(1-\lambda^2)+2=0$$

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