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The locus of the point of intersection of tangents drawn at the extremities of a normal chord of the parabola $y^2=-8x$ is a curve having asymptote at $x$=...?

Let one end of the normal chord be $(-2t^2,-4t)$,then the other parametric coordinate of the chord is $t'=-t-1/t$ and therefore the other end of chord is $(-2(t+1/t)^2,4(t+1/t))$.The point of intersection of the tangents is $(-2tt',-2(t+t'))$.Eliminating $t,t'$ I got the locus to be $x^2=2(16/y^2+2)$.However I couldnot proceed to find out the asymptote to it.Please help me in this regard.Thanks.

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