The foci of the hyperbola frac{x^2}{16} - fra | vedclass

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Question

(A) The given equation is $\frac{x^2}{16} - \frac{(y - 2)^2}{9} = 1$.Comparing this with the standard form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2}

Vedclass · Accepted Answer

$(5, 2), (-5, 2)$

Vedclass · Answer

(A) The given equation is $\frac{x^2}{16} - \frac{(y - 2)^2}{9} = 1$.Comparing this with the standard form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$,we have $h = 0$,$k = 2$,$a^2 = 16$,and $b^2 = 9$.Thus,$a = 4$ and $b = 3$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.The foci are given by $(h \pm ae, k)$.Substituting the values,we get $(0 \pm 4 	imes \frac{5}{4}, 2) = (\pm 5, 2)$.Therefore,the foci are $(5, 2)$ and $(-5, 2)$.

The foci of the hyperbola $\frac{x^2}{16} - \frac{(y - 2)^2}{9} = 1$ are

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