Answer:
the equation of the hyperbola centered at the origin is
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]
[tex]\frac{x^2}{\left(5\sqrt{5}\right)^2}-\frac{y^2}{\left(5i\right)^2}=1[/tex]
Step-by-step explanation:
For a right-left facing hyperbola, the Foci (focus points) are defined:
[tex]\left(h+c,\:k\right),\:\left(h-c,\:k\right)[/tex]
where [tex]c=\sqrt{a^2+b^2}[/tex] is the distance form the center (h, k) to a focus.
So, the standard equation will be
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]
As the directrix is the line is
[tex]x=\frac{a^2}{c}[/tex]
As c = 10, and x = 125/10
So,
[tex]\:x\times c=a^2[/tex]
[tex]\sqrt{x\times \:c}=a[/tex]
[tex]\sqrt{\frac{125}{10}\times \:10}=a[/tex]
[tex]5\sqrt{5}=a[/tex]
As
[tex]a^2+b^2=c^2[/tex]
[tex]b^2=c^2-a^2[/tex]
[tex]b=\sqrt{c^2-a^2}[/tex]
[tex]b=\sqrt{\left(10\right)^2-\left(5\sqrt{5}\right)^2}[/tex]
[tex]b=\sqrt{100-125}[/tex]
[tex]b=\sqrt{-25}[/tex]
[tex]b=\sqrt{-1}\sqrt{25}[/tex]
[tex]\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i[/tex]
[tex]b=\sqrt{25}i[/tex]
[tex]b=5i[/tex]
Therefore, the equation of the hyperbola centered at the origin is
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]
[tex]\frac{x^2}{\left(5\sqrt{5}\right)^2}-\frac{y^2}{\left(5i\right)^2}=1[/tex]
Keywords: hyperbola, directrex, foci, equation of hyperbola
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