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(B) The equation of the rectangular hyperbola is $x^2 - y^2 = 1$,where $a=1$ and $b=1$. For a point $P(	heta)$ on the hyperbola,the coordinates are $

Vedclass · Accepted Answer

$57$

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(B) The equation of the rectangular hyperbola is $x^2 - y^2 = 1$,where $a=1$ and $b=1$. For a point $P(	heta)$ on the hyperbola,the coordinates are $(\sec 	heta, 	an 	heta)$. At $	heta_1 = \frac{\pi}{4}$,the point $P$ is $(\sec \frac{\pi}{4}, 	an \frac{\pi}{4}) = (\sqrt{2}, 1)$. The slope of the tangent at $(\sec 	heta, 	an 	heta)$ is $\frac{dy}{dx} = \frac{\sec 	heta 	an 	heta}{\sec^2 	heta} = \sin 	heta$. At $P$,the slope of the tangent is $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$. The slope of the normal at $P$ is $-\frac{1}{	ext{slope of tangent}} = -\sqrt{2}$. The equation of the normal at $P(\sqrt{2}, 1)$ is $y - 1 = -\sqrt{2}(x - \sqrt{2})$,which simplifies to $y = -\sqrt{2}x + 3$. Substituting this into the hyperbola equation $x^2 - y^2 = 1$: $x^2 - (-\sqrt{2}x + 3)^2 = 1$ $x^2 - (2x^2 - 6\sqrt{2}x + 9) = 1$ $-x^2 + 6\sqrt{2}x - 10 = 0$ $x^2 - 6\sqrt{2}x + 10 = 0$. Since $x_1 = \sqrt{2}$ is a root,the product of roots $x_1 x_2 = 10$,so $x_2 = \frac{10}{\sqrt{2}} = 5\sqrt{2}$. For point $Q$,$x_2 = \sec 	heta_2 = 5\sqrt{2}$. Then $	an^2 	heta_2 = \sec^2 	heta_2 - 1 = (5\sqrt{2})^2 - 1 = 50 - 1 = 49$,so $	an 	heta_2 = 7$ (since $y_2 = -\sqrt{2}(5\sqrt{2}) + 3 = -10 + 3 = -7$,but $	an 	heta$ is defined by the point coordinates). Thus,$\sec^2 	heta_2 + 	an 	heta_2 = 50 + 7 = 57$.

If the normal to the rectangular hyperbola $x^2-y^2=1$ at the point $P$ with parameter $\theta_1 = \frac{\pi}{4}$ meets the curve again at $Q$ with parameter $\theta_2$,then find the value of $\sec^2 \theta_2 + \tan \theta_2$.

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