<s>[INST] Find the equation of the locus of the intersection of the lines below Find the equation of the locus of the intersection of the lines below
$y=mx+\sqrt{m^2+2}$
$y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}$

By graphing, I have got an ellipse as locus : $x^2+\dfrac{y^2}{2}=1$.
The given lines form tangent and normal to above ellipse.
Is there any other nice way to eliminate $m$ w/o using graphing ?
I have tried eliminating $m$ by solving the intersection point, but it's looking very messy. Thanks!
 [/INST] Notice, 
Solving the given equations of the straight lines: $y=mx+\sqrt{m^2+2}$ & $y=-\frac{1}{m}x+\sqrt{\frac{1}{m^2+2}}$,
as follows
$$mx+\sqrt{m^2+2}=-\frac{1}{m}x+\sqrt{\frac{1}{m^2+2}}$$
$$mx+\frac{x}{m}=\frac{1}{\sqrt{m^2+2}}-\sqrt{m^2+2}$$
$$\frac{m^2+1}{m}x=\frac{1-m^2-2}{\sqrt{m^2+2}}\implies x=\frac{-m}{\sqrt{m^2+2}}$$
substituting value of $x$ in first equation we get 
$$y=m\frac{-m}{\sqrt{m^2+2}}+\sqrt{m^2+2}\implies y=\frac{2}{\sqrt{m^2+2}}$$
If the intersection point is $(h, k)$ then we have 
$$h=\frac{-m}{\sqrt{m^2+2}}\tag 1$$$$ \ k=\frac{2}{\sqrt{m^2+2}}\tag 2$$
dividing (1) by (2), we get 
$$\frac{h}{k}=-\frac{m}{2}\implies m=-2\frac{h}{k}$$
Now, squaring (1), we get 
$$h^2=\frac{m^2}{m^2+2}$$ setting the value of $m$ we get
$$h^2=\frac{\left(-\frac{2h}{k}\right)^2}{\left(-\frac{2h}{k}\right)^2+2}$$
$$4\frac{h^2}{k^2}+2=\frac{4}{k^2}$$
$$h^2+\frac{k^2}{2}=1$$
Substituting $h=x$ & $k=y$, the locus of the intersection point is given as 
$$\color{red}{x^2+\frac{y^2}{2}=1}$$ 
Above equation represents an ellipse. 
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