If the eccentricity of the standard hyperbola | vedclass

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(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Since it passes through $(4, 6)$,we have $\frac{16}{a^2} - \frac{36}{

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$2x - y - 2 = 0$

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(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Since it passes through $(4, 6)$,we have $\frac{16}{a^2} - \frac{36}{b^2} = 1 \dots (i)$.Given eccentricity $e = 2$,we have $e^2 = 1 + \frac{b^2}{a^2}$ $\Rightarrow 4 = 1 + \frac{b^2}{a^2}$ $\Rightarrow b^2 = 3a^2 \dots (ii)$.Substituting $(ii)$ into $(i)$,we get $\frac{16}{a^2} - \frac{36}{3a^2} = 1$ $\Rightarrow \frac{16 - 12}{a^2} = 1$ $\Rightarrow a^2 = 4$.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 $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.$\frac{4x}{4} - \frac{6y}{12} = 1$ $\Rightarrow x - \frac{y}{2} = 1$ $\Rightarrow 2x - y = 2$ $\Rightarrow 2x - y - 2 = 0$.

If the eccentricity of the standard hyperbola passing through the point $(4, 6)$ is $2$,then the equation of the tangent to the hyperbola at $(4, 6)$ is

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