If the normal to the rectangular hyperbola $x^{2} - y^{2} = 4$ at a point $P$ meets the coordinates

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If the normal to the rectangular hyperbola $x^{2} - y^{2} = 4$ at a point $P$ meets the coordinates axes in $Q$ and $R$ and $O$ is the centre of the hyperbola. Then:

Accepted Answer

Sol. (a, b, c, d)Equation of normal at $P$$$\frac{x}{sec	heta} + \frac{y}{tan	heta} = 4$$$$\begin{matrix} \Rightarrow \ Q & \ \equiv (4sec	heta,0),R \equiv (0,4tan	heta) \ 	herefore\ OP & \ = PQ = PR = 2\sqrt{\sec^{2}	heta + 	an^{2}	heta} \ QR & \ = 4\sqrt{\sec^{2}	heta + 	an^{2}	heta} \end{matrix}$$

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