How do I simplify this:
$$0.15\sqrt{(x-2171)^2+y^2 }+551.6≤0.15\sqrt{(x+2171)^2+y^2}$$
to become a hyperbola: $$\frac{x^2}{1839^2}+\frac{y^2}{1155^2} ≥1$$
Would appreciate if anyone could point me in the right direction. THANKS!
How do I simplify this:
$$0.15\sqrt{(x-2171)^2+y^2 }+551.6≤0.15\sqrt{(x+2171)^2+y^2}$$
to become a hyperbola: $$\frac{x^2}{1839^2}+\frac{y^2}{1155^2} ≥1$$
Would appreciate if anyone could point me in the right direction. THANKS!
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Hint: with $\,a=2171\,$ and $\,b=551.6 \,/\, 0.15\,$ the inequality can be written as:
$$ \sqrt{(x+a)^2+y^2} - \sqrt{(x-a)^2+y^2} \ge b $$
Squaring both sides and expanding:
$$ \require{cancel} x^2+\cancel{2ax}+a^2+y^2+x^2-\cancel{2ax}+a^2+y^2 - 2 \sqrt{\left((x+a)^2+y^2\right)\left((x-a)^2+y^2\right)} \ge b^2 $$
Rearranging and squaring again:
$$ (2x^2+2y^2+2a^2-b^2)^2 \ge 4\left( (x^2-a^2)^2+y^4+2y^2(x^2+a^2) \right) $$
It can be seen by inspection that the terms in $x^4,y^4$ and $x^2 \cdot y^2$ cancel out between the two sides, so what remains is a conic inequality which can be explicitly written out by expanding the terms.