At which of the following points is the tange | vedclass

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(B) The equation of the hyperbola is $x^2 - y^2 = 3$. Differentiating with respect to $x$,we get $2x - 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{

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$(2, -1)$

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(B) The equation of the hyperbola is $x^2 - y^2 = 3$. Differentiating with respect to $x$,we get $2x - 2y \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = \frac{x}{y}$. The slope of the tangent at any point $(x_1, y_1)$ is $m = \frac{x_1}{y_1}$. The given line is $2x + y + 8 = 0$,which can be written as $y = -2x - 8$. The slope of this line is $-2$. Since the tangent is parallel to the line,the slopes must be equal: $\frac{x_1}{y_1} = -2$,so $x_1 = -2y_1$. Substituting this into the hyperbola equation: $(-2y_1)^2 - y_1^2 = 3$. $4y_1^2 - y_1^2 = 3$,which gives $3y_1^2 = 3$,so $y_1^2 = 1$,meaning $y_1 = \pm 1$. If $y_1 = 1$,then $x_1 = -2(1) = -2$. Point is $(-2, 1)$. If $y_1 = -1$,then $x_1 = -2(-1) = 2$. Point is $(2, -1)$. Comparing with the options,$(2, -1)$ is present.

At which of the following points is the tangent to the hyperbola $x^2 - y^2 = 3$ parallel to the line $2x + y + 8 = 0$?

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