Let C be the centre of the hyperbola frac{x^2 | vedclass

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(A) Let $P$ be $(a \sec 	heta, b 	an 	heta)$.The tangent at $P$ is $\frac{x \sec 	heta}{a} - \frac{y 	an 	heta}{b} = 1$.It meets $bx - ay = 0$ (

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$a^2 + b^2$

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(A) Let $P$ be $(a \sec 	heta, b 	an 	heta)$.The tangent at $P$ is $\frac{x \sec 	heta}{a} - \frac{y 	an 	heta}{b} = 1$.It meets $bx - ay = 0$ (i.e.,$\frac{x}{a} = \frac{y}{b}$) at point $Q$.Substituting $y = \frac{bx}{a}$ into the tangent equation: $\frac{x \sec 	heta}{a} - \frac{bx 	an 	heta}{ab} = 1 \implies \frac{x}{a}(\sec 	heta - 	an 	heta) = 1 \implies x = \frac{a}{\sec 	heta - 	an 	heta}, y = \frac{b}{\sec 	heta - 	an 	heta}$.Thus,$Q = (\frac{a}{\sec 	heta - 	an 	heta}, \frac{b}{\sec 	heta - 	an 	heta})$.It meets $bx + ay = 0$ (i.e.,$\frac{x}{a} = -\frac{y}{b}$) at point $R$.Substituting $y = -\frac{bx}{a}$ into the tangent equation: $\frac{x \sec 	heta}{a} + \frac{bx 	an 	heta}{ab} = 1 \implies \frac{x}{a}(\sec 	heta + 	an 	heta) = 1 \implies x = \frac{a}{\sec 	heta + 	an 	heta}, y = -\frac{b}{\sec 	heta + 	an 	heta}$.Thus,$R = (\frac{a}{\sec 	heta + 	an 	heta}, -\frac{b}{\sec 	heta + 	an 	heta})$.$CQ = \sqrt{(\frac{a}{\sec 	heta - 	an 	heta})^2 + (\frac{b}{\sec 	heta - 	an 	heta})^2} = \frac{\sqrt{a^2 + b^2}}{\sec 	heta - 	an 	heta}$.$CR = \sqrt{(\frac{a}{\sec 	heta + 	an 	heta})^2 + (-\frac{b}{\sec 	heta + 	an 	heta})^2} = \frac{\sqrt{a^2 + b^2}}{\sec 	heta + 	an 	heta}$.$CQ \cdot CR = \frac{a^2 + b^2}{(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta)} = \frac{a^2 + b^2}{\sec^2 	heta - 	an^2 	heta} = a^2 + b^2$ (since $\sec^2 	heta - 	an^2 	heta = 1$).

Let $C$ be the centre of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The tangent at any point $P$ on this hyperbola meets the straight lines $bx - ay = 0$ and $bx + ay = 0$ at points $Q$ and $R$ respectively. Then $CQ \cdot CR = $

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