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(D) The asymptotes of the rectangular hyperbola $x^2 - y^2 = a^2$ are $y = x$ and $y = -x$.Let $P$ be a point $(a \sec 	heta, a 	an 	heta)$ on the

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a hyperbola

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(D) The asymptotes of the rectangular hyperbola $x^2 - y^2 = a^2$ are $y = x$ and $y = -x$.Let $P$ be a point $(a \sec 	heta, a 	an 	heta)$ on the hyperbola.Consider the asymptote $x - y = 0$.The perpendicular distance from $P(a \sec 	heta, a 	an 	heta)$ to the line $x - y = 0$ is $PN = \frac{|a \sec 	heta - a 	an 	heta|}{\sqrt{1^2 + (-1)^2}} = \frac{a}{\sqrt{2}} |\sec 	heta - 	an 	heta|$.The coordinates of $N$ can be found by the projection of $P$ onto the line $x - y = 0$.The line passing through $P$ perpendicular to $x - y = 0$ is $x + y = a(\sec 	heta + 	an 	heta)$.Solving $x - y = 0$ and $x + y = a(\sec 	heta + 	an 	heta)$,we get $N = (\frac{a}{2}(\sec 	heta + 	an 	heta), \frac{a}{2}(\sec 	heta + 	an 	heta))$.Let the midpoint of $PN$ be $(h, k)$. Then $h = \frac{a \sec 	heta + \frac{a}{2}(\sec 	heta + 	an 	heta)}{2} = \frac{a}{4}(3 \sec 	heta + 	an 	heta)$ and $k = \frac{a 	an 	heta + \frac{a}{2}(\sec 	heta + 	an 	heta)}{2} = \frac{a}{4}(\sec 	heta + 3 	an 	heta)$.Calculating $h^2 - k^2 = \frac{a^2}{16} [(3 \sec 	heta + 	an 	heta)^2 - (\sec 	heta + 3 	an 	heta)^2] = \frac{a^2}{16} [8 \sec^2 	heta - 8 	an^2 	heta] = \frac{a^2}{2} (\sec^2 	heta - 	an^2 	heta) = \frac{a^2}{2}$.Thus,the locus is $x^2 - y^2 = \frac{a^2}{2}$,which is a hyperbola.

If $PN$ is the perpendicular from a point $P$ on the rectangular hyperbola $x^2 - y^2 = a^2$ to any of its asymptotes,then the locus of the midpoint of $PN$ is:

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