If the eccentricity of the hyperbola frac{x^2 | vedclass

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(D) Given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with eccentricity $e = 2$. We know that $e^2 = 1 + \frac{b^2}{a^2}$,so $4 = 1 + \frac{

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$2x - y - 2 = 0$

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(D) Given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with eccentricity $e = 2$. We know that $e^2 = 1 + \frac{b^2}{a^2}$,so $4 = 1 + \frac{b^2}{a^2}$,which implies $\frac{b^2}{a^2} = 3$,or $b^2 = 3a^2$. The hyperbola passes through $(4, 6)$,so $\frac{16}{a^2} - \frac{36}{b^2} = 1$. Substituting $b^2 = 3a^2$,we get $\frac{16}{a^2} - \frac{36}{3a^2} = 1$,which simplifies to $\frac{16}{a^2} - \frac{12}{a^2} = 1$,so $\frac{4}{a^2} = 1$,giving $a^2 = 4$. Then $b^2 = 3(4) = 12$. The equation of the hyperbola is $\frac{x^2}{4} - \frac{y^2}{12} = 1$. The equation of the tangent at $(x_1, y_1) = (4, 6)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$. Substituting the values,we get $\frac{4x}{4} - \frac{6y}{12} = 1$,which simplifies to $x - \frac{y}{2} = 1$. Multiplying by $2$,we get $2x - y = 2$,or $2x - y - 2 = 0$.

If the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passing through the point $(4,6)$ is $2$,then the equation of the tangent to this hyperbola at $(4,6)$ is

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