Let a, r, s, t be nonzero real numbers. Let P | vedclass

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(D,B) $1.$ Since $PQ$ is a focal chord,$t \cdot t' = -1$,so $t' = -\frac{1}{t}$.The slope of $PK$ is $m_{PK} = \frac{2at - 0}{at^2 - 2a} = \frac{2at}{

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$(D, B)$

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(D,B) $1.$ Since $PQ$ is a focal chord,$t \cdot t' = -1$,so $t' = -\frac{1}{t}$.The slope of $PK$ is $m_{PK} = \frac{2at - 0}{at^2 - 2a} = \frac{2at}{a(t^2-2)} = \frac{2t}{t^2-2}$.The slope of $QR$ is $m_{QR} = \frac{2ar - 2at'}{ar^2 - at'^2} = \frac{2a(r-t')}{a(r-t')(r+t')} = \frac{2}{r+t'}$.Since $QR \parallel PK$,$m_{QR} = m_{PK} \implies \frac{2}{r+t'} = \frac{2t}{t^2-2}$.$r+t' = \frac{t^2-2}{t} = t - \frac{2}{t}$.Substituting $t' = -\frac{1}{t}$,we get $r - \frac{1}{t} = t - \frac{2}{t} \implies r = t - \frac{1}{t} = \frac{t^2-1}{t}$.Thus,the correct option for $1$ is $(D)$.$2.$ The tangent at $P(at^2, 2at)$ is $ty = x + at^2$.The normal at $S(as^2, 2as)$ is $y = -sx + 2as + as^3$,or $y + sx = 2as + as^3$.Given $st = 1$,so $s = \frac{1}{t}$.The normal equation becomes $y + \frac{1}{t}x = 2a(\frac{1}{t}) + a(\frac{1}{t^3}) = \frac{2at^2+a}{t^3}$.$ty + x = \frac{2at^2+a}{t^2}$.We have the system:$ty - x = at^2$$ty + x = \frac{2at^2+a}{t^2}$Adding the two equations: $2ty = at^2 + \frac{2at^2+a}{t^2} = \frac{at^4 + 2at^2 + a}{t^2} = \frac{a(t^2+1)^2}{t^2}$.$y = \frac{a(t^2+1)^2}{2t^3}$.Thus,the correct option for $2$ is $(B)$.

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