Let the tangent to the parabola y^2=12x at th | vedclass

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(A) Since the point $P(3, \alpha)$ lies on the parabola $y^2=12x$,we have $\alpha^2 = 12(3) = 36$,so $\alpha = \pm 6$.The slope of the tangent to $y^2

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$116$

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(A) Since the point $P(3, \alpha)$ lies on the parabola $y^2=12x$,we have $\alpha^2 = 12(3) = 36$,so $\alpha = \pm 6$.The slope of the tangent to $y^2=12x$ at $(3, \alpha)$ is given by $\frac{dy}{dx} = \frac{6}{y}$. At $(3, \alpha)$,the slope is $m_1 = \frac{6}{\alpha}$.The line $2x+2y=3$ has a slope $m_2 = -1$. Since the tangent is perpendicular to this line,$m_1 	imes m_2 = -1$,which implies $\frac{6}{\alpha} 	imes (-1) = -1$,so $\alpha = 6$.Now,the hyperbola equation is $6^2x^2 - 9y^2 = 9(6^2)$,which simplifies to $36x^2 - 9y^2 = 324$,or $\frac{x^2}{9} - \frac{y^2}{36} = 1$.The point $Q$ on the hyperbola is $(\alpha-1, \alpha+2) = (5, 8)$.The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $(x_0, y_0)$ is $\frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2+b^2$. Here $a^2=9, b^2=36, x_0=5, y_0=8$.So,$\frac{9x}{5} + \frac{36y}{8} = 9+36 = 45$,which simplifies to $\frac{9x}{5} + \frac{9y}{2} = 45$. Dividing by $9$,we get $\frac{x}{5} + \frac{y}{2} = 5$,or $2x + 5y - 50 = 0$.The distance $d$ of the point $(6, -4)$ from the line $2x + 5y - 50 = 0$ is $d = \frac{|2(6) + 5(-4) - 50|}{\sqrt{2^2 + 5^2}} = \frac{|12 - 20 - 50|}{\sqrt{29}} = \frac{|-58|}{\sqrt{29}} = \frac{58}{\sqrt{29}}$.The square of the distance is $d^2 = \frac{58^2}{29} = \frac{3364}{29} = 116$.

Let the tangent to the parabola $y^2=12x$ at the point $(3, \alpha)$ be perpendicular to the line $2x+2y=3$. Then the square of the distance of the point $(6, -4)$ from the normal to the hyperbola $\alpha^2x^2-9y^2=9\alpha^2$ at its point $(\alpha-1, \alpha+2)$ is equal to $........$.

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