Find the equation of the tangent to the hyper | vedclass

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(A) The given equation of the hyperbola is $\frac{x^2}{3} - \frac{y^2}{2} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2

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$y = x + 1$

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(A) The given equation of the hyperbola is $\frac{x^2}{3} - \frac{y^2}{2} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 3$ and $b^2 = 2$.Since the tangent is equally inclined to the axes,its slope $m$ must be $	an(45^{\circ}) = 1$ or $	an(135^{\circ}) = -1$.Taking $m = 1$,the equation of the tangent to the hyperbola $y = mx \pm \sqrt{a^2m^2 - b^2}$ is given by:$y = 1 \cdot x \pm \sqrt{3(1)^2 - 2}$$y = x \pm \sqrt{3 - 2}$$y = x \pm 1$.Thus,the possible equations are $y = x + 1$ or $y = x - 1$. Given the options,$y = x + 1$ is the correct choice.

Find the equation of the tangent to the hyperbola $\frac{x^2}{3} - \frac{y^2}{2} = 1$ which is equally inclined to the axes.

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