Question

write the equation in standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0).

Answer 1

Standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0) is $\frac{x^{2} }{144}-\frac{y^{2} }{25}=1$

Hyperbola is a plane curve generated by a point so moving that the difference of the distances from two fixed points is a constant.

Given,

The Vertices of the hyperbola =  (-12,0) and (12,0)

Foci = (-13,0) and (13,0)

a=12

ac=13

c=$\frac{13}{12}$

We know,

c=$\sqrt{1+\frac{b^{2} }{a^{2} } }$

$c^{2}=1+\frac{b^{2} }{a^{2} } \\(\frac{13}{12} )^{2}=1+\frac{b^{2} }{144} \\\frac{169}{144}=1+ \frac{b^{2} }{144} \\\frac{b^{2} }{144} =\frac{25}{144}\\ b^{2}=25$

The equation of the hyperbola is

$\frac{x^{2} }{a^{2} }-\frac{y^{2} }{b^{2} }=1\\ \frac{x^{2} }{144 }-\frac{y^{2} }{25 }=1$

Hence, the standard form for the hyperbola with vertices ( – 12,0) and (12,0), and foci ( – 13,0) and (13,0) is $\frac{x^{2} }{144}-\frac{y^{2} }{25}=1$

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