The equation of the tangent parallel to y - x | vedclass

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(A) The given hyperbola is $\frac{x^2}{3} - \frac{y^2}{2} = 1$. Comparing this with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 3$ and $b^2

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$x - y - 1 = 0$

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(A) The given hyperbola is $\frac{x^2}{3} - \frac{y^2}{2} = 1$. Comparing this with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 3$ and $b^2 = 2$. The equation of the tangent parallel to $y - x + 5 = 0$ (which is $y = x - 5$,slope $m = 1$) is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$. Substituting $m = 1, a^2 = 3, b^2 = 2$: $y = 1 \cdot x \pm \sqrt{3(1)^2 - 2}$ $y = x \pm \sqrt{3 - 2}$ $y = x \pm 1$ This gives two equations: $x - y + 1 = 0$ and $x - y - 1 = 0$. Comparing with the given options,$x - y - 1 = 0$ is the correct answer.

The equation of the tangent parallel to $y - x + 5 = 0$ drawn to the hyperbola $\frac{x^2}{3} - \frac{y^2}{2} = 1$ is

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