The foci of a hyperbola are (pm 3, 0) and the | vedclass

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(A) Given the foci are $(\pm ae, 0) = (\pm 3, 0)$,we have $ae = 3$,so $a^{2}e^{2} = 9$.For a hyperbola,$b^{2} = a^{2}e^{2} - a^{2}$,thus $a^{2} + b^{2

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$4x^{2} - 5y^{2} = 20$

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(A) Given the foci are $(\pm ae, 0) = (\pm 3, 0)$,we have $ae = 3$,so $a^{2}e^{2} = 9$.For a hyperbola,$b^{2} = a^{2}e^{2} - a^{2}$,thus $a^{2} + b^{2} = a^{2}e^{2} = 9$ (Equation $i$).The equation of the tangent $y = mx + c$ is $y = -2x + 4$,where $m = -2$ and $c = 4$.The condition for tangency to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $c^{2} = a^{2}m^{2} - b^{2}$.Substituting the values: $4^{2} = a^{2}(-2)^{2} - b^{2} \Rightarrow 4a^{2} - b^{2} = 16$ (Equation $ii$).Adding Equation $(i)$ and $(ii)$: $(a^{2} + b^{2}) + (4a^{2} - b^{2}) = 9 + 16$ $\Rightarrow 5a^{2} = 25$ $\Rightarrow a^{2} = 5$.Substituting $a^{2} = 5$ into Equation $(i)$: $5 + b^{2} = 9 \Rightarrow b^{2} = 4$.The equation of the hyperbola is $\frac{x^{2}}{5} - \frac{y^{2}}{4} = 1$,which simplifies to $4x^{2} - 5y^{2} = 20$.

The foci of a hyperbola are $(\pm 3, 0)$ and the equation of a tangent is $2x + y - 4 = 0$. Find the equation of the hyperbola.

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