Consider the hyperbola H : x^2-y^2=1 and a ci | vedclass

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(B) The tangent to the hyperbola $H: x^2 - y^2 = 1$ at $P(x_1, y_1)$ is $xx_1 - yy_1 = 1$. Setting $y = 0$,we find the intersection with the $x$-axis

Vedclass · Accepted Answer

$(A, B, D)$

Vedclass · Answer

(B) The tangent to the hyperbola $H: x^2 - y^2 = 1$ at $P(x_1, y_1)$ is $xx_1 - yy_1 = 1$. Setting $y = 0$,we find the intersection with the $x$-axis at $M(\frac{1}{x_1}, 0)$. The slope of the tangent at $P$ is $\frac{x_1}{y_1}$,so the slope of the normal is $-\frac{y_1}{x_1}$. The normal passes through the center $N(x_2, 0)$ and $P(x_1, y_1)$,so $-\frac{y_1}{x_1} = \frac{y_1 - 0}{x_1 - x_2}$. This simplifies to $x_2 - x_1 = x_1$,so $x_2 = 2x_1$. Thus $N = (2x_1, 0)$. The centroid $(l, m)$ of $	riangle PMN$ with vertices $P(x_1, y_1)$,$M(\frac{1}{x_1}, 0)$,and $N(2x_1, 0)$ is given by $l = \frac{x_1 + \frac{1}{x_1} + 2x_1}{3} = \frac{3x_1 + \frac{1}{x_1}}{3} = x_1 + \frac{1}{3x_1}$ and $m = \frac{y_1 + 0 + 0}{3} = \frac{y_1}{3}$. Calculating the derivatives: $\frac{dl}{dx_1} = \frac{d}{dx_1}(x_1 + \frac{1}{3x_1}) = 1 - \frac{1}{3x_1^2}$. This matches option $(A)$. Since $y_1^2 = x_1^2 - 1$,we have $y_1 = \sqrt{x_1^2 - 1}$. $\frac{dm}{dx_1} = \frac{1}{3} \frac{dy_1}{dx_1} = \frac{1}{3} \cdot \frac{x_1}{\sqrt{x_1^2 - 1}}$. This matches option $(B)$. $\frac{dm}{dy_1} = \frac{d}{dy_1}(\frac{y_1}{3}) = \frac{1}{3}$. This matches option $(D)$. Thus,the correct options are $(A, B, D)$.

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