The equation of the tangents to the hyperbola | vedclass

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Question

(B) The given hyperbola is $3x^2 - 4y^2 = 12$,which can be written as $\frac{x^2}{4} - \frac{y^2}{3} = 1$.Here,$a^2 = 4$ and $b^2 = 3$.The equation of

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$y - x = \pm 1$

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(B) The given hyperbola is $3x^2 - 4y^2 = 12$,which can be written as $\frac{x^2}{4} - \frac{y^2}{3} = 1$.Here,$a^2 = 4$ and $b^2 = 3$.The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.Since the tangent cuts equal intercepts from the axes,its equation is of the form $\frac{x}{c} + \frac{y}{c} = 1$ or $\frac{x}{c} - \frac{y}{c} = 1$,which implies the slope $m = \pm 1$.For $m = 1$,the tangent is $y = x \pm \sqrt{4(1)^2 - 3} = x \pm \sqrt{1} = x \pm 1$,which gives $y - x = \pm 1$.For $m = -1$,the tangent is $y = -x \pm \sqrt{4(-1)^2 - 3} = -x \pm 1$,which gives $y + x = \pm 1$.However,checking the options provided,the correct equation is $y - x = \pm 1$.

The equation of the tangents to the hyperbola $3x^2 - 4y^2 = 12$ which cut equal intercepts from the axes are:

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