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Exercise: Hyperbola · Q27

Q.Find the coordinates of the foci, the vertices, the eccentricity and the equations of the asymptotes of the hyperbola 16y2−9x2=14416y^2-9x^2=144.

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Divide 16y2−9x2=14416y^2-9x^2=144 by 144144: y29−x216=1\dfrac{y^2}{9}-\dfrac{x^2}{16}=1. Since the positive term is y2/9y^2/9, the transverse axis is along yy: a2=9⇒a=3a^2=9\Rightarrow a=3, b2=16⇒b=4b^2=16\Rightarrow b=4.

c2=a2+b2=9+16=25⇒c=5.c^2=a^2+b^2=9+16=25\Rightarrow c=5.

Foci =(0,±5)=(0,\pm5), vertices =(0,±3)=(0,\pm3), e=c/a=5/3e=c/a=5/3, asymptotes y=±abx=±34xy=\pm\dfrac{a}{b}x=\pm\dfrac34x.

✓Final answer

Foci (0,±5)(0,\pm5), vertices (0,±3)(0,\pm3), e=5/3e=5/3

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