Question

Find the Vertices, Foci and Asymptotes of this conic: 16y^2 -x^2 = 1

Answer 1

\(\bf {\color{blue}{ \cfrac{(y-k)^2}{a^2}-\cfrac{(x-h)^2}{b^2}=1}}
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16y^2 -x^2 = 1\implies 4^2y^2 -x^2 = 1\implies \cfrac{1}{4^{-2}}\cdot y^2 -x^2 = 1
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\cfrac{y^2}{\frac{1}{16}}-\cfrac{x^2}{1}=1\implies \cfrac{(y-0)^2}{\frac{1}{16}}-\cfrac{(x-0)^2}{1}=1
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\cfrac{(y-0)^2}{\left(\sqrt{\frac{1}{16}}\right)^2}-\cfrac{(x-0)^2}{1^2}=1\\
-----------------------------\\
vertices\implies (h\quad ,\quad k\pm a )\qquad foci\implies (h\quad ,\quad k\pm \sqrt{a^2+b^2})
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asymptotes\implies k\pm \cfrac{a}{b}(x-h)\)