Question

EN LA IMAGEN ESTA LA ECUACION HIPERBOLA!!
Deducir la ecuación de la hipérbola:

Answer 1

$\text{Si }c\ \textgreater \ a\ \textgreater \ b\text{ entonces: } c^2=a^2+b^2\\ \\ \sqrt{(x-c)^2+y^2}-\sqrt{(x+c)^2+y^2}=\pm2a\\ \\ \left(\sqrt{(x-c)^2+y^2}-\sqrt{(x+c)^2+y^2}\right)^2=(\pm2a)^2\\ \\ (x-c)^2+y^2+(x+c)^2+y^2-2\sqrt{[(x-c)^2+y^2][(x+c)^2+y^2]}=4a^2$

$(x-c)^2+(x+c)^2+2y^2-2\sqrt{[(x^2+c^2+y^2)-2xc][(x^2+c^2+y^2)+2xc]}=4a^2\\ \\ 2x^2+2c^2+2y^2-2\sqrt{[(x^2+c^2+y^2)-2xc][(x^2+c^2+y^2)+2xc]}=4a^2\\ \\ \sqrt{(x^2+c^2+y^2)^2-(2xc)^2}=2a^2-x^2-y^2-c^2\\ \\ \sqrt{(x^2+y^2+c^2)^2-(2xc)^2}=2a^2-(x^2+y^2+c^2)$

$(x^2+y^2+c^2)^2-(2xc)^2=[2a^2-(x^2+y^2+c^2)]^2\\ \\ (x^2+y^2+c^2)^2-(2xc)^2=4a^4+(x^2+y^2+c^2)^2-4a^2(x^2+y^2+c^2)\\ \\ -(2xc)^2=4a^4-4a^2(x^2+y^2+c^2)\\ \\ -x^2c^2=a^4-a^2x^2-a^2(y^2+c^2)$

$a^4+(c^2-a^2)x^2-a^2(y^2+a^2+b^2)=0\\ \\ b^2x^2-a^2(y^2+b^2)=0\\ \\ b^2x^2-a^2y^2=a^2b^2\\ \\ \boxed{\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1}$