If a hyperbola passes through the point P(sqr | vedclass

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(A) The standard equation of a hyperbola with foci at $(\pm ae, 0)$ is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Given foci are $(\pm 2, 0)$, so $ae = 2

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$y = x\sqrt{6} - \sqrt{3}$

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(A) The standard equation of a hyperbola with foci at $(\pm ae, 0)$ is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Given foci are $(\pm 2, 0)$, so $ae = 2$. Also, $b^2 = a^2(e^2 - 1) = a^2e^2 - a^2 = 4 - a^2$.Since the hyperbola passes through $P(\sqrt{2}, \sqrt{3})$, we have $\frac{(\sqrt{2})^2}{a^2} - \frac{(\sqrt{3})^2}{b^2} = 1$, which simplifies to $\frac{2}{a^2} - \frac{3}{4 - a^2} = 1$.Let $a^2 = u$. Then $\frac{2}{u} - \frac{3}{4 - u} = 1$ $\Rightarrow 2(4 - u) - 3u = u(4 - u)$ $\Rightarrow 8 - 2u - 3u = 4u - u^2$.Rearranging gives $u^2 - 9u + 8 = 0$, so $(u - 8)(u - 1) = 0$. Since $a^2 Then $b^2 = 4 - 1 = 3$. The equation is $\frac{x^2}{1} - \frac{y^2}{3} = 1$.The tangent at $P(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.Substituting $P(\sqrt{2}, \sqrt{3})$, $a^2=1$, $b^2=3$: $\frac{x\sqrt{2}}{1} - \frac{y\sqrt{3}}{3} = 1 \Rightarrow x\sqrt{2} - \frac{y}{\sqrt{3}} = 1$.Multiplying by $\sqrt{3}$, we get $x\sqrt{6} - y = \sqrt{3}$, or $y = x\sqrt{6} - \sqrt{3}$.

If a hyperbola passes through the point $P(\sqrt{2}, \sqrt{3})$ and has foci at $(\pm 2, 0)$, then the tangent to this hyperbola at $P$ is:

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